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Keep running int errors while trying to ask my code to run a y/n while loop [duplicate]

I attempted to create a calculator, but I can not get it to work because I don't know how to get user input.
How can I get the user input in Java?

One of the simplest ways is to use a Scanner object as follows:
import java.util.Scanner;
Scanner reader = new Scanner(System.in); // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();

You can use any of the following options based on the requirements.
Scanner class
import java.util.Scanner;
//...
Scanner scan = new Scanner(System.in);
String s = scan.next();
int i = scan.nextInt();
BufferedReader and InputStreamReader classes
import java.io.BufferedReader;
import java.io.InputStreamReader;
//...
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
int i = Integer.parseInt(s);
DataInputStream class
import java.io.DataInputStream;
//...
DataInputStream dis = new DataInputStream(System.in);
int i = dis.readInt();
The readLine method from the DataInputStream class has been deprecated. To get String value, you should use the previous solution with BufferedReader
Console class
import java.io.Console;
//...
Console console = System.console();
String s = console.readLine();
int i = Integer.parseInt(console.readLine());
Apparently, this method does not work well in some IDEs.

You can use the Scanner class or the Console class
Console console = System.console();
String input = console.readLine("Enter input:");

You can get user input using BufferedReader.
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String accStr;
System.out.println("Enter your Account number: ");
accStr = br.readLine();
It will store a String value in accStr so you have to parse it to an int using Integer.parseInt.
int accInt = Integer.parseInt(accStr);

Here is how you can get the keyboard inputs:
Scanner scanner = new Scanner (System.in);
System.out.print("Enter your name");
String name = scanner.next(); // Get what the user types.

The best two options are BufferedReader and Scanner.
The most widely used method is Scanner and I personally prefer it because of its simplicity and easy implementation, as well as its powerful utility to parse text into primitive data.
Advantages of Using Scanner
Easy to use the Scanner class
Easy input of numbers (int, short, byte, float, long and double)
Exceptions are unchecked which is more convenient. It is up to the programmer to be civilized, and specify or catch the exceptions.
Is able to read lines, white spaces, and regex-delimited tokens
Advantages of BufferedInputStream
BufferedInputStream is about reading in blocks of data rather than a single byte at a time
Can read chars, char arrays, and lines
Throws checked exceptions
Fast performance
Synchronized (you cannot share Scanner between threads)
Overall each input method has different purposes.
If you are inputting large amount of data BufferedReader might be
better for you
If you are inputting lots of numbers Scanner does automatic parsing
which is very convenient
For more basic uses I would recommend the Scanner because it is easier to use and easier to write programs with. Here is a quick example of how to create a Scanner. I will provide a comprehensive example below of how to use the Scanner
Scanner scanner = new Scanner (System.in); // create scanner
System.out.print("Enter your name"); // prompt user
name = scanner.next(); // get user input
(For more info about BufferedReader see How to use a BufferedReader and see Reading lines of Chars)
java.util.Scanner
import java.util.InputMismatchException; // import the exception catching class
import java.util.Scanner; // import the scanner class
public class RunScanner {
// main method which will run your program
public static void main(String args[]) {
// create your new scanner
// Note: since scanner is opened to "System.in" closing it will close "System.in".
// Do not close scanner until you no longer want to use it at all.
Scanner scanner = new Scanner(System.in);
// PROMPT THE USER
// Note: when using scanner it is recommended to prompt the user with "System.out.print" or "System.out.println"
System.out.println("Please enter a number");
// use "try" to catch invalid inputs
try {
// get integer with "nextInt()"
int n = scanner.nextInt();
System.out.println("Please enter a decimal"); // PROMPT
// get decimal with "nextFloat()"
float f = scanner.nextFloat();
System.out.println("Please enter a word"); // PROMPT
// get single word with "next()"
String s = scanner.next();
// ---- Note: Scanner.nextInt() does not consume a nextLine character /n
// ---- In order to read a new line we first need to clear the current nextLine by reading it:
scanner.nextLine();
// ----
System.out.println("Please enter a line"); // PROMPT
// get line with "nextLine()"
String l = scanner.nextLine();
// do something with the input
System.out.println("The number entered was: " + n);
System.out.println("The decimal entered was: " + f);
System.out.println("The word entered was: " + s);
System.out.println("The line entered was: " + l);
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input entered. Please enter the specified input");
}
scanner.close(); // close the scanner so it doesn't leak
}
}
Note: Other classes such as Console and DataInputStream are also viable alternatives.
Console has some powerful features such as ability to read passwords, however, is not available in all IDE's (such as Eclipse). The reason this occurs is because Eclipse runs your application as a background process and not as a top-level process with a system console. Here is a link to a useful example on how to implement the Console class.
DataInputStream is primarily used for reading input as a primitive datatype, from an underlying input stream, in a machine-independent way. DataInputStream is usually used for reading binary data. It also provides convenience methods for reading certain data types. For example, it has a method to read a UTF String which can contain any number of lines within them.
However, it is a more complicated class and harder to implement so not recommended for beginners. Here is a link to a useful example how to implement a DataInputStream.

You can make a simple program to ask for user's name and print what ever the reply use inputs.
Or ask user to enter two numbers and you can add, multiply, subtract, or divide those numbers and print the answers for user inputs just like a behavior of a calculator.
So there you need Scanner class. You have to import java.util.Scanner; and in the code you need to use
Scanner input = new Scanner(System.in);
Input is a variable name.
Scanner input = new Scanner(System.in);
System.out.println("Please enter your name : ");
s = input.next(); // getting a String value
System.out.println("Please enter your age : ");
i = input.nextInt(); // getting an integer
System.out.println("Please enter your salary : ");
d = input.nextDouble(); // getting a double
See how this differs: input.next();, i = input.nextInt();, d = input.nextDouble();
According to a String, int and a double varies same way for the rest. Don't forget the import statement at the top of your code.
Also see the blog post "Scanner class and getting User Inputs".

To read a line or a string, you can use a BufferedReader object combined with an InputStreamReader one as follows:
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(System.in));
String inputLine = bufferReader.readLine();

Here, the program asks the user to enter a number. After that, the program prints the digits of the number and the sum of the digits.
import java.util.Scanner;
public class PrintNumber {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int num = 0;
int sum = 0;
System.out.println(
"Please enter a number to show its digits");
num = scan.nextInt();
System.out.println(
"Here are the digits and the sum of the digits");
while (num > 0) {
System.out.println("==>" + num % 10);
sum += num % 10;
num = num / 10;
}
System.out.println("Sum is " + sum);
}
}

Here is your program from the question using java.util.Scanner:
import java.util.Scanner;
public class Example {
public static void main(String[] args) {
int input = 0;
System.out.println("The super insano calculator");
System.out.println("enter the corrosponding number:");
Scanner reader3 = new Scanner(System.in);
System.out.println(
"1. Add | 2. Subtract | 3. Divide | 4. Multiply");
input = reader3.nextInt();
int a = 0, b = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Enter the first number");
// get user input for a
a = reader.nextInt();
Scanner reader1 = new Scanner(System.in);
System.out.println("Enter the scend number");
// get user input for b
b = reader1.nextInt();
switch (input){
case 1: System.out.println(a + " + " + b + " = " + add(a, b));
break;
case 2: System.out.println(a + " - " + b + " = " + subtract(a, b));
break;
case 3: System.out.println(a + " / " + b + " = " + divide(a, b));
break;
case 4: System.out.println(a + " * " + b + " = " + multiply(a, b));
break;
default: System.out.println("your input is invalid!");
break;
}
}
static int add(int lhs, int rhs) { return lhs + rhs; }
static int subtract(int lhs, int rhs) { return lhs - rhs; }
static int divide(int lhs, int rhs) { return lhs / rhs; }
static int multiply(int lhs, int rhs) { return lhs * rhs; }
}

Scanner input = new Scanner(System.in);
String inputval = input.next();

Scanner input=new Scanner(System.in);
int integer=input.nextInt();
String string=input.next();
long longInteger=input.nextLong();

Just one extra detail. If you don't want to risk a memory/resource leak, you should close the scanner stream when you are finished:
myScanner.close();
Note that java 1.7 and later catch this as a compile warning (don't ask how I know that :-)

Here is a more developed version of the accepted answer that addresses two common needs:
Collecting user input repeatedly until an exit value has been entered
Dealing with invalid input values (non-integers in this example)
Code
package inputTest;
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");
boolean done = false;
while (!done) {
System.out.print("Enter an integer: ");
try {
int n = reader.nextInt();
if (n == 0) {
done = true;
}
else {
// do something with the input
System.out.println("\tThe number entered was: " + n);
}
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.
}
}
System.out.println("Exiting...");
reader.close();
}
}
Example
Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...
Note that without nextLine(), the bad input will trigger the same exception repeatedly in an infinite loop. You might want to use next() instead depending on the circumstance, but know that input like this has spaces will generate multiple exceptions.

import java.util.Scanner;
class Daytwo{
public static void main(String[] args){
System.out.println("HelloWorld");
Scanner reader = new Scanner(System.in);
System.out.println("Enter the number ");
int n = reader.nextInt();
System.out.println("You entered " + n);
}
}

Add throws IOException beside main(), then
DataInputStream input = new DataInputStream(System.in);
System.out.print("Enter your name");
String name = input.readLine();

It is very simple to get input in java, all you have to do is:
import java.util.Scanner;
class GetInputFromUser
{
public static void main(String args[])
{
int a;
float b;
String s;
Scanner in = new Scanner(System.in);
System.out.println("Enter a string");
s = in.nextLine();
System.out.println("You entered string " + s);
System.out.println("Enter an integer");
a = in.nextInt();
System.out.println("You entered integer " + a);
System.out.println("Enter a float");
b = in.nextFloat();
System.out.println("You entered float " + b);
}
}

import java.util.Scanner;
public class Myapplication{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int a;
System.out.println("enter:");
a = in.nextInt();
System.out.println("Number is= " + a);
}
}

You can get user input like this using a BufferedReader:
InputStreamReader inp = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(inp);
// you will need to import these things.
This is how you apply them
String name = br.readline();
So when the user types in his name into the console, "String name" will store that information.
If it is a number you want to store, the code will look like this:
int x = Integer.parseInt(br.readLine());
Hop this helps!

Can be something like this...
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
System.out.println("Enter a number: ");
int i = reader.nextInt();
for (int j = 0; j < i; j++)
System.out.println("I love java");
}

You can get the user input using Scanner. You can use the proper input validation using proper methods for different data types like next() for String or nextInt() for Integer.
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
//reads the input until it reaches the space
System.out.println("Enter a string: ");
String str = scanner.next();
System.out.println("str = " + str);
//reads until the end of line
String aLine = scanner.nextLine();
//reads the integer
System.out.println("Enter an integer num: ");
int num = scanner.nextInt();
System.out.println("num = " + num);
//reads the double value
System.out.println("Enter a double: ");
double aDouble = scanner.nextDouble();
System.out.println("double = " + aDouble);
//reads the float value, long value, boolean value, byte and short
double aFloat = scanner.nextFloat();
long aLong = scanner.nextLong();
boolean aBoolean = scanner.nextBoolean();
byte aByte = scanner.nextByte();
short aShort = scanner.nextShort();
scanner.close();

import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Welcome to the best program in the world! ");
while (true) {
System.out.print("Enter a query: ");
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
if (s.equals("q")) {
System.out.println("The program is ending now ....");
break;
} else {
System.out.println("The program is running...");
}
}
}
}

This is a simple code that uses the System.in.read() function. This code just writes out whatever was typed. You can get rid of the while loop if you just want to take input once, and you could store answers in a character array if you so choose.
package main;
import java.io.IOException;
public class Root
{
public static void main(String[] args)
{
new Root();
}
public Root()
{
while(true)
{
try
{
for(int y = 0; y < System.in.available(); ++y)
{
System.out.print((char)System.in.read());
}
}
catch(IOException ex)
{
ex.printStackTrace(System.out);
break;
}
}
}
}

I like the following:
public String readLine(String tPromptString) {
byte[] tBuffer = new byte[256];
int tPos = 0;
System.out.print(tPromptString);
while(true) {
byte tNextByte = readByte();
if(tNextByte == 10) {
return new String(tBuffer, 0, tPos);
}
if(tNextByte != 13) {
tBuffer[tPos] = tNextByte;
++tPos;
}
}
}
and for example, I would do:
String name = this.readLine("What is your name?")

Keyboard entry using Scanner is possible, as others have posted. But in these highly graphic times it is pointless making a calculator without a graphical user interface (GUI).
In modern Java this means using a JavaFX drag-and-drop tool like Scene Builder to lay out a GUI that resembles a calculator's console.
Note that using Scene Builder is intuitively easy and demands no additional Java skill for its event handlers that what you already may have.
For user input, you should have a wide TextField at the top of the GUI console.
This is where the user enters the numbers that they want to perform functions on.
Below the TextField, you would have an array of function buttons doing basic (i.e. add/subtract/multiply/divide and memory/recall/clear) functions.
Once the GUI is lain out, you can then add the 'controller' references that link each button function to its Java implementation, e.g a call to method in your project's controller class.
This video is a bit old but still shows how easy Scene Builder is to use.

The most simple way to get user input would be to use Scanner. Here's an example of how it's supposed to be used:
import java.util.Scanner;
public class main {
public static void main(String[]args) {
Scanner sc=new Scanner(System.in);
int a;
String b;
System.out.println("Type an integer here: ");
a=sc.nextInt();
System.out.println("Type anything here:");
b=sc.nextLine();
The line of code import java.util.Scanner; tells the program that the programmer will be using user inputs in their code. Like it says, it imports the scanner utility. Scanner sc=new Scanner(System.in); tells the program to start the user inputs. After you do that, you must make a string or integer without a value, then put those in the line a=sc.nextInt(); or a=sc.nextLine();. This gives the variables the value of the user inputs. Then you can use it in your code. Hope this helps.

Using JOptionPane you can achieve it.
Int a =JOptionPane.showInputDialog(null,"Enter number:");

import java.util.Scanner;
public class userinput {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Name : ");
String name = input.next();
System.out.print("Last Name : ");
String lname = input.next();
System.out.print("Age : ");
byte age = input.nextByte();
System.out.println(" " );
System.out.println(" " );
System.out.println("Firt Name: " + name);
System.out.println("Last Name: " + lname);
System.out.println(" Age: " + age);
}
}

class ex1 {
public static void main(String args[]){
int a, b, c;
a = Integer.parseInt(args[0]);
b = Integer.parseInt(args[1]);
c = a + b;
System.out.println("c = " + c);
}
}
// Output
javac ex1.java
java ex1 10 20
c = 30

How do I read and print numbers in Java? [duplicate]

What class can I use for reading an integer variable in Java?

You can use java.util.Scanner (API):
import java.util.Scanner;
//...
Scanner in = new Scanner(System.in);
int num = in.nextInt();
It can also tokenize input with regular expression, etc. The API has examples and there are many others in this site (e.g. How do I keep a scanner from throwing exceptions when the wrong type is entered?).

If you are using Java 6, you can use the following oneliner to read an integer from console:
int n = Integer.parseInt(System.console().readLine());

Here I am providing 2 examples to read integer value from the standard input
Example 1
import java.util.Scanner;
public class Maxof2
{
public static void main(String args[])
{
//taking value as command line argument.
Scanner in = new Scanner(System.in);
System.out.printf("Enter i Value: ");
int i = in.nextInt();
System.out.printf("Enter j Value: ");
int j = in.nextInt();
if(i > j)
System.out.println(i+"i is greater than "+j);
else
System.out.println(j+" is greater than "+i);
}
}
Example 2
public class ReadandWritewhateveryoutype
{
public static void main(String args[]) throws java.lang.Exception
{
System.out.printf("This Program is used to Read and Write what ever you type \nType quit to Exit at any Moment\n\n");
java.io.BufferedReader r = new java.io.BufferedReader (new java.io.InputStreamReader (System.in));
String hi;
while (!(hi=r.readLine()).startsWith("quit"))System.out.printf("\nYou have typed: %s \n",hi);
}
}
I prefer the First Example, it's easy and quite understandable.
You can compile and run the JAVA programs online at this website: http://ideone.com

Check this one:
public static void main(String[] args) {
String input = null;
int number = 0;
try {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
input = bufferedReader.readLine();
number = Integer.parseInt(input);
} catch (NumberFormatException ex) {
System.out.println("Not a number !");
} catch (IOException e) {
e.printStackTrace();
}
}

Second answer above is the most simple one.
int n = Integer.parseInt(System.console().readLine());
The question is "How to read from standard input".
A console is a device typically associated to the keyboard and display from which a program is launched.
You may wish to test if no Java console device is available, e.g. Java VM not started from a command line or the standard input and output streams are redirected.
Console cons;
if ((cons = System.console()) == null) {
System.err.println("Unable to obtain console");
...
}
Using console is a simple way to input numbers. Combined with parseInt()/Double() etc.
s = cons.readLine("Enter a int: ");
int i = Integer.parseInt(s);
s = cons.readLine("Enter a double: ");
double d = Double.parseDouble(s);

check this one:
import java.io.*;
public class UserInputInteger
{
public static void main(String args[])throws IOException
{
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int number;
System.out.println("Enter the number");
number = Integer.parseInt(in.readLine());
}
}

This causes headaches so I updated a solution that will run using the most common hardware and software tools available to users in December 2014. Please note that the JDK/SDK/JRE/Netbeans and their subsequent classes, template libraries compilers, editors and debuggerz are free.
This program was tested with Java v8 u25. It was written and built using
Netbeans IDE 8.0.2, JDK 1.8, OS is win8.1 (apologies) and browser is Chrome (double-apologies)
- meant to assist UNIX-cmd-line OG's deal with modern GUI-Web-based IDEs
at ZERO COST - because information (and IDEs) should always be free.
By Tapper7. For Everyone.
code block:
package modchk; //Netbeans requirement.
import java.util.Scanner;
//import java.io.*; is not needed Netbeans automatically includes it.
public class Modchk {
public static void main(String[] args){
int input1;
int input2;
//Explicity define the purpose of the .exe to user:
System.out.println("Modchk by Tapper7. Tests IOStream and basic bool modulo fxn.\n"
+ "Commented and coded for C/C++ programmers new to Java\n");
//create an object that reads integers:
Scanner Cin = new Scanner(System.in);
//the following will throw() if you don't do you what it tells you or if
//int entered == ArrayIndex-out-of-bounds for your system. +-~2.1e9
System.out.println("Enter an integer wiseguy: ");
input1 = Cin.nextInt(); //this command emulates "cin >> input1;"
//I test like Ernie Banks played hardball: "Let's play two!"
System.out.println("Enter another integer...anyday now: ");
input2 = Cin.nextInt();
//debug the scanner and istream:
System.out.println("the 1st N entered by the user was " + input1);
System.out.println("the 2nd N entered by the user was " + input2);
//"do maths" on vars to make sure they are of use to me:
System.out.println("modchk for " + input1);
if(2 % input1 == 0){
System.out.print(input1 + " is even\n"); //<---same output effect as *.println
}else{
System.out.println(input1 + " is odd");
}//endif input1
//one mo' 'gain (as in istream dbg chk above)
System.out.println("modchk for " + input2);
if(2 % input2 == 0){
System.out.print(input2 + " is even\n");
}else{
System.out.println(input2 + " is odd");
}//endif input2
}//end main
}//end Modchk

Word Count: user defined [duplicate]

I attempted to create a calculator, but I can not get it to work because I don't know how to get user input.
How can I get the user input in Java?

One of the simplest ways is to use a Scanner object as follows:
import java.util.Scanner;
Scanner reader = new Scanner(System.in); // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();

You can use any of the following options based on the requirements.
Scanner class
import java.util.Scanner;
//...
Scanner scan = new Scanner(System.in);
String s = scan.next();
int i = scan.nextInt();
BufferedReader and InputStreamReader classes
import java.io.BufferedReader;
import java.io.InputStreamReader;
//...
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
int i = Integer.parseInt(s);
DataInputStream class
import java.io.DataInputStream;
//...
DataInputStream dis = new DataInputStream(System.in);
int i = dis.readInt();
The readLine method from the DataInputStream class has been deprecated. To get String value, you should use the previous solution with BufferedReader
Console class
import java.io.Console;
//...
Console console = System.console();
String s = console.readLine();
int i = Integer.parseInt(console.readLine());
Apparently, this method does not work well in some IDEs.

You can use the Scanner class or the Console class
Console console = System.console();
String input = console.readLine("Enter input:");

You can get user input using BufferedReader.
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String accStr;
System.out.println("Enter your Account number: ");
accStr = br.readLine();
It will store a String value in accStr so you have to parse it to an int using Integer.parseInt.
int accInt = Integer.parseInt(accStr);

Here is how you can get the keyboard inputs:
Scanner scanner = new Scanner (System.in);
System.out.print("Enter your name");
String name = scanner.next(); // Get what the user types.

The best two options are BufferedReader and Scanner.
The most widely used method is Scanner and I personally prefer it because of its simplicity and easy implementation, as well as its powerful utility to parse text into primitive data.
Advantages of Using Scanner
Easy to use the Scanner class
Easy input of numbers (int, short, byte, float, long and double)
Exceptions are unchecked which is more convenient. It is up to the programmer to be civilized, and specify or catch the exceptions.
Is able to read lines, white spaces, and regex-delimited tokens
Advantages of BufferedInputStream
BufferedInputStream is about reading in blocks of data rather than a single byte at a time
Can read chars, char arrays, and lines
Throws checked exceptions
Fast performance
Synchronized (you cannot share Scanner between threads)
Overall each input method has different purposes.
If you are inputting large amount of data BufferedReader might be
better for you
If you are inputting lots of numbers Scanner does automatic parsing
which is very convenient
For more basic uses I would recommend the Scanner because it is easier to use and easier to write programs with. Here is a quick example of how to create a Scanner. I will provide a comprehensive example below of how to use the Scanner
Scanner scanner = new Scanner (System.in); // create scanner
System.out.print("Enter your name"); // prompt user
name = scanner.next(); // get user input
(For more info about BufferedReader see How to use a BufferedReader and see Reading lines of Chars)
java.util.Scanner
import java.util.InputMismatchException; // import the exception catching class
import java.util.Scanner; // import the scanner class
public class RunScanner {
// main method which will run your program
public static void main(String args[]) {
// create your new scanner
// Note: since scanner is opened to "System.in" closing it will close "System.in".
// Do not close scanner until you no longer want to use it at all.
Scanner scanner = new Scanner(System.in);
// PROMPT THE USER
// Note: when using scanner it is recommended to prompt the user with "System.out.print" or "System.out.println"
System.out.println("Please enter a number");
// use "try" to catch invalid inputs
try {
// get integer with "nextInt()"
int n = scanner.nextInt();
System.out.println("Please enter a decimal"); // PROMPT
// get decimal with "nextFloat()"
float f = scanner.nextFloat();
System.out.println("Please enter a word"); // PROMPT
// get single word with "next()"
String s = scanner.next();
// ---- Note: Scanner.nextInt() does not consume a nextLine character /n
// ---- In order to read a new line we first need to clear the current nextLine by reading it:
scanner.nextLine();
// ----
System.out.println("Please enter a line"); // PROMPT
// get line with "nextLine()"
String l = scanner.nextLine();
// do something with the input
System.out.println("The number entered was: " + n);
System.out.println("The decimal entered was: " + f);
System.out.println("The word entered was: " + s);
System.out.println("The line entered was: " + l);
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input entered. Please enter the specified input");
}
scanner.close(); // close the scanner so it doesn't leak
}
}
Note: Other classes such as Console and DataInputStream are also viable alternatives.
Console has some powerful features such as ability to read passwords, however, is not available in all IDE's (such as Eclipse). The reason this occurs is because Eclipse runs your application as a background process and not as a top-level process with a system console. Here is a link to a useful example on how to implement the Console class.
DataInputStream is primarily used for reading input as a primitive datatype, from an underlying input stream, in a machine-independent way. DataInputStream is usually used for reading binary data. It also provides convenience methods for reading certain data types. For example, it has a method to read a UTF String which can contain any number of lines within them.
However, it is a more complicated class and harder to implement so not recommended for beginners. Here is a link to a useful example how to implement a DataInputStream.

You can make a simple program to ask for user's name and print what ever the reply use inputs.
Or ask user to enter two numbers and you can add, multiply, subtract, or divide those numbers and print the answers for user inputs just like a behavior of a calculator.
So there you need Scanner class. You have to import java.util.Scanner; and in the code you need to use
Scanner input = new Scanner(System.in);
Input is a variable name.
Scanner input = new Scanner(System.in);
System.out.println("Please enter your name : ");
s = input.next(); // getting a String value
System.out.println("Please enter your age : ");
i = input.nextInt(); // getting an integer
System.out.println("Please enter your salary : ");
d = input.nextDouble(); // getting a double
See how this differs: input.next();, i = input.nextInt();, d = input.nextDouble();
According to a String, int and a double varies same way for the rest. Don't forget the import statement at the top of your code.
Also see the blog post "Scanner class and getting User Inputs".

To read a line or a string, you can use a BufferedReader object combined with an InputStreamReader one as follows:
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(System.in));
String inputLine = bufferReader.readLine();

Here, the program asks the user to enter a number. After that, the program prints the digits of the number and the sum of the digits.
import java.util.Scanner;
public class PrintNumber {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int num = 0;
int sum = 0;
System.out.println(
"Please enter a number to show its digits");
num = scan.nextInt();
System.out.println(
"Here are the digits and the sum of the digits");
while (num > 0) {
System.out.println("==>" + num % 10);
sum += num % 10;
num = num / 10;
}
System.out.println("Sum is " + sum);
}
}

Here is your program from the question using java.util.Scanner:
import java.util.Scanner;
public class Example {
public static void main(String[] args) {
int input = 0;
System.out.println("The super insano calculator");
System.out.println("enter the corrosponding number:");
Scanner reader3 = new Scanner(System.in);
System.out.println(
"1. Add | 2. Subtract | 3. Divide | 4. Multiply");
input = reader3.nextInt();
int a = 0, b = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Enter the first number");
// get user input for a
a = reader.nextInt();
Scanner reader1 = new Scanner(System.in);
System.out.println("Enter the scend number");
// get user input for b
b = reader1.nextInt();
switch (input){
case 1: System.out.println(a + " + " + b + " = " + add(a, b));
break;
case 2: System.out.println(a + " - " + b + " = " + subtract(a, b));
break;
case 3: System.out.println(a + " / " + b + " = " + divide(a, b));
break;
case 4: System.out.println(a + " * " + b + " = " + multiply(a, b));
break;
default: System.out.println("your input is invalid!");
break;
}
}
static int add(int lhs, int rhs) { return lhs + rhs; }
static int subtract(int lhs, int rhs) { return lhs - rhs; }
static int divide(int lhs, int rhs) { return lhs / rhs; }
static int multiply(int lhs, int rhs) { return lhs * rhs; }
}

Scanner input = new Scanner(System.in);
String inputval = input.next();

Scanner input=new Scanner(System.in);
int integer=input.nextInt();
String string=input.next();
long longInteger=input.nextLong();

Just one extra detail. If you don't want to risk a memory/resource leak, you should close the scanner stream when you are finished:
myScanner.close();
Note that java 1.7 and later catch this as a compile warning (don't ask how I know that :-)

Here is a more developed version of the accepted answer that addresses two common needs:
Collecting user input repeatedly until an exit value has been entered
Dealing with invalid input values (non-integers in this example)
Code
package inputTest;
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");
boolean done = false;
while (!done) {
System.out.print("Enter an integer: ");
try {
int n = reader.nextInt();
if (n == 0) {
done = true;
}
else {
// do something with the input
System.out.println("\tThe number entered was: " + n);
}
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.
}
}
System.out.println("Exiting...");
reader.close();
}
}
Example
Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...
Note that without nextLine(), the bad input will trigger the same exception repeatedly in an infinite loop. You might want to use next() instead depending on the circumstance, but know that input like this has spaces will generate multiple exceptions.

import java.util.Scanner;
class Daytwo{
public static void main(String[] args){
System.out.println("HelloWorld");
Scanner reader = new Scanner(System.in);
System.out.println("Enter the number ");
int n = reader.nextInt();
System.out.println("You entered " + n);
}
}

Add throws IOException beside main(), then
DataInputStream input = new DataInputStream(System.in);
System.out.print("Enter your name");
String name = input.readLine();

It is very simple to get input in java, all you have to do is:
import java.util.Scanner;
class GetInputFromUser
{
public static void main(String args[])
{
int a;
float b;
String s;
Scanner in = new Scanner(System.in);
System.out.println("Enter a string");
s = in.nextLine();
System.out.println("You entered string " + s);
System.out.println("Enter an integer");
a = in.nextInt();
System.out.println("You entered integer " + a);
System.out.println("Enter a float");
b = in.nextFloat();
System.out.println("You entered float " + b);
}
}

import java.util.Scanner;
public class Myapplication{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int a;
System.out.println("enter:");
a = in.nextInt();
System.out.println("Number is= " + a);
}
}

You can get user input like this using a BufferedReader:
InputStreamReader inp = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(inp);
// you will need to import these things.
This is how you apply them
String name = br.readline();
So when the user types in his name into the console, "String name" will store that information.
If it is a number you want to store, the code will look like this:
int x = Integer.parseInt(br.readLine());
Hop this helps!

Can be something like this...
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
System.out.println("Enter a number: ");
int i = reader.nextInt();
for (int j = 0; j < i; j++)
System.out.println("I love java");
}

You can get the user input using Scanner. You can use the proper input validation using proper methods for different data types like next() for String or nextInt() for Integer.
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
//reads the input until it reaches the space
System.out.println("Enter a string: ");
String str = scanner.next();
System.out.println("str = " + str);
//reads until the end of line
String aLine = scanner.nextLine();
//reads the integer
System.out.println("Enter an integer num: ");
int num = scanner.nextInt();
System.out.println("num = " + num);
//reads the double value
System.out.println("Enter a double: ");
double aDouble = scanner.nextDouble();
System.out.println("double = " + aDouble);
//reads the float value, long value, boolean value, byte and short
double aFloat = scanner.nextFloat();
long aLong = scanner.nextLong();
boolean aBoolean = scanner.nextBoolean();
byte aByte = scanner.nextByte();
short aShort = scanner.nextShort();
scanner.close();

import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Welcome to the best program in the world! ");
while (true) {
System.out.print("Enter a query: ");
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
if (s.equals("q")) {
System.out.println("The program is ending now ....");
break;
} else {
System.out.println("The program is running...");
}
}
}
}

This is a simple code that uses the System.in.read() function. This code just writes out whatever was typed. You can get rid of the while loop if you just want to take input once, and you could store answers in a character array if you so choose.
package main;
import java.io.IOException;
public class Root
{
public static void main(String[] args)
{
new Root();
}
public Root()
{
while(true)
{
try
{
for(int y = 0; y < System.in.available(); ++y)
{
System.out.print((char)System.in.read());
}
}
catch(IOException ex)
{
ex.printStackTrace(System.out);
break;
}
}
}
}

I like the following:
public String readLine(String tPromptString) {
byte[] tBuffer = new byte[256];
int tPos = 0;
System.out.print(tPromptString);
while(true) {
byte tNextByte = readByte();
if(tNextByte == 10) {
return new String(tBuffer, 0, tPos);
}
if(tNextByte != 13) {
tBuffer[tPos] = tNextByte;
++tPos;
}
}
}
and for example, I would do:
String name = this.readLine("What is your name?")

Keyboard entry using Scanner is possible, as others have posted. But in these highly graphic times it is pointless making a calculator without a graphical user interface (GUI).
In modern Java this means using a JavaFX drag-and-drop tool like Scene Builder to lay out a GUI that resembles a calculator's console.
Note that using Scene Builder is intuitively easy and demands no additional Java skill for its event handlers that what you already may have.
For user input, you should have a wide TextField at the top of the GUI console.
This is where the user enters the numbers that they want to perform functions on.
Below the TextField, you would have an array of function buttons doing basic (i.e. add/subtract/multiply/divide and memory/recall/clear) functions.
Once the GUI is lain out, you can then add the 'controller' references that link each button function to its Java implementation, e.g a call to method in your project's controller class.
This video is a bit old but still shows how easy Scene Builder is to use.

The most simple way to get user input would be to use Scanner. Here's an example of how it's supposed to be used:
import java.util.Scanner;
public class main {
public static void main(String[]args) {
Scanner sc=new Scanner(System.in);
int a;
String b;
System.out.println("Type an integer here: ");
a=sc.nextInt();
System.out.println("Type anything here:");
b=sc.nextLine();
The line of code import java.util.Scanner; tells the program that the programmer will be using user inputs in their code. Like it says, it imports the scanner utility. Scanner sc=new Scanner(System.in); tells the program to start the user inputs. After you do that, you must make a string or integer without a value, then put those in the line a=sc.nextInt(); or a=sc.nextLine();. This gives the variables the value of the user inputs. Then you can use it in your code. Hope this helps.

Using JOptionPane you can achieve it.
Int a =JOptionPane.showInputDialog(null,"Enter number:");

import java.util.Scanner;
public class userinput {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Name : ");
String name = input.next();
System.out.print("Last Name : ");
String lname = input.next();
System.out.print("Age : ");
byte age = input.nextByte();
System.out.println(" " );
System.out.println(" " );
System.out.println("Firt Name: " + name);
System.out.println("Last Name: " + lname);
System.out.println(" Age: " + age);
}
}

class ex1 {
public static void main(String args[]){
int a, b, c;
a = Integer.parseInt(args[0]);
b = Integer.parseInt(args[1]);
c = a + b;
System.out.println("c = " + c);
}
}
// Output
javac ex1.java
java ex1 10 20
c = 30

How to get the user input in Java?

I attempted to create a calculator, but I can not get it to work because I don't know how to get user input.
How can I get the user input in Java?

One of the simplest ways is to use a Scanner object as follows:
import java.util.Scanner;
Scanner reader = new Scanner(System.in); // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();

You can use any of the following options based on the requirements.
Scanner class
import java.util.Scanner;
//...
Scanner scan = new Scanner(System.in);
String s = scan.next();
int i = scan.nextInt();
BufferedReader and InputStreamReader classes
import java.io.BufferedReader;
import java.io.InputStreamReader;
//...
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
int i = Integer.parseInt(s);
DataInputStream class
import java.io.DataInputStream;
//...
DataInputStream dis = new DataInputStream(System.in);
int i = dis.readInt();
The readLine method from the DataInputStream class has been deprecated. To get String value, you should use the previous solution with BufferedReader
Console class
import java.io.Console;
//...
Console console = System.console();
String s = console.readLine();
int i = Integer.parseInt(console.readLine());
Apparently, this method does not work well in some IDEs.

You can use the Scanner class or the Console class
Console console = System.console();
String input = console.readLine("Enter input:");

You can get user input using BufferedReader.
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String accStr;
System.out.println("Enter your Account number: ");
accStr = br.readLine();
It will store a String value in accStr so you have to parse it to an int using Integer.parseInt.
int accInt = Integer.parseInt(accStr);

Here is how you can get the keyboard inputs:
Scanner scanner = new Scanner (System.in);
System.out.print("Enter your name");
String name = scanner.next(); // Get what the user types.

The best two options are BufferedReader and Scanner.
The most widely used method is Scanner and I personally prefer it because of its simplicity and easy implementation, as well as its powerful utility to parse text into primitive data.
Advantages of Using Scanner
Easy to use the Scanner class
Easy input of numbers (int, short, byte, float, long and double)
Exceptions are unchecked which is more convenient. It is up to the programmer to be civilized, and specify or catch the exceptions.
Is able to read lines, white spaces, and regex-delimited tokens
Advantages of BufferedInputStream
BufferedInputStream is about reading in blocks of data rather than a single byte at a time
Can read chars, char arrays, and lines
Throws checked exceptions
Fast performance
Synchronized (you cannot share Scanner between threads)
Overall each input method has different purposes.
If you are inputting large amount of data BufferedReader might be
better for you
If you are inputting lots of numbers Scanner does automatic parsing
which is very convenient
For more basic uses I would recommend the Scanner because it is easier to use and easier to write programs with. Here is a quick example of how to create a Scanner. I will provide a comprehensive example below of how to use the Scanner
Scanner scanner = new Scanner (System.in); // create scanner
System.out.print("Enter your name"); // prompt user
name = scanner.next(); // get user input
(For more info about BufferedReader see How to use a BufferedReader and see Reading lines of Chars)
java.util.Scanner
import java.util.InputMismatchException; // import the exception catching class
import java.util.Scanner; // import the scanner class
public class RunScanner {
// main method which will run your program
public static void main(String args[]) {
// create your new scanner
// Note: since scanner is opened to "System.in" closing it will close "System.in".
// Do not close scanner until you no longer want to use it at all.
Scanner scanner = new Scanner(System.in);
// PROMPT THE USER
// Note: when using scanner it is recommended to prompt the user with "System.out.print" or "System.out.println"
System.out.println("Please enter a number");
// use "try" to catch invalid inputs
try {
// get integer with "nextInt()"
int n = scanner.nextInt();
System.out.println("Please enter a decimal"); // PROMPT
// get decimal with "nextFloat()"
float f = scanner.nextFloat();
System.out.println("Please enter a word"); // PROMPT
// get single word with "next()"
String s = scanner.next();
// ---- Note: Scanner.nextInt() does not consume a nextLine character /n
// ---- In order to read a new line we first need to clear the current nextLine by reading it:
scanner.nextLine();
// ----
System.out.println("Please enter a line"); // PROMPT
// get line with "nextLine()"
String l = scanner.nextLine();
// do something with the input
System.out.println("The number entered was: " + n);
System.out.println("The decimal entered was: " + f);
System.out.println("The word entered was: " + s);
System.out.println("The line entered was: " + l);
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input entered. Please enter the specified input");
}
scanner.close(); // close the scanner so it doesn't leak
}
}
Note: Other classes such as Console and DataInputStream are also viable alternatives.
Console has some powerful features such as ability to read passwords, however, is not available in all IDE's (such as Eclipse). The reason this occurs is because Eclipse runs your application as a background process and not as a top-level process with a system console. Here is a link to a useful example on how to implement the Console class.
DataInputStream is primarily used for reading input as a primitive datatype, from an underlying input stream, in a machine-independent way. DataInputStream is usually used for reading binary data. It also provides convenience methods for reading certain data types. For example, it has a method to read a UTF String which can contain any number of lines within them.
However, it is a more complicated class and harder to implement so not recommended for beginners. Here is a link to a useful example how to implement a DataInputStream.

You can make a simple program to ask for user's name and print what ever the reply use inputs.
Or ask user to enter two numbers and you can add, multiply, subtract, or divide those numbers and print the answers for user inputs just like a behavior of a calculator.
So there you need Scanner class. You have to import java.util.Scanner; and in the code you need to use
Scanner input = new Scanner(System.in);
Input is a variable name.
Scanner input = new Scanner(System.in);
System.out.println("Please enter your name : ");
s = input.next(); // getting a String value
System.out.println("Please enter your age : ");
i = input.nextInt(); // getting an integer
System.out.println("Please enter your salary : ");
d = input.nextDouble(); // getting a double
See how this differs: input.next();, i = input.nextInt();, d = input.nextDouble();
According to a String, int and a double varies same way for the rest. Don't forget the import statement at the top of your code.
Also see the blog post "Scanner class and getting User Inputs".

To read a line or a string, you can use a BufferedReader object combined with an InputStreamReader one as follows:
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(System.in));
String inputLine = bufferReader.readLine();

Here, the program asks the user to enter a number. After that, the program prints the digits of the number and the sum of the digits.
import java.util.Scanner;
public class PrintNumber {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int num = 0;
int sum = 0;
System.out.println(
"Please enter a number to show its digits");
num = scan.nextInt();
System.out.println(
"Here are the digits and the sum of the digits");
while (num > 0) {
System.out.println("==>" + num % 10);
sum += num % 10;
num = num / 10;
}
System.out.println("Sum is " + sum);
}
}

Here is your program from the question using java.util.Scanner:
import java.util.Scanner;
public class Example {
public static void main(String[] args) {
int input = 0;
System.out.println("The super insano calculator");
System.out.println("enter the corrosponding number:");
Scanner reader3 = new Scanner(System.in);
System.out.println(
"1. Add | 2. Subtract | 3. Divide | 4. Multiply");
input = reader3.nextInt();
int a = 0, b = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Enter the first number");
// get user input for a
a = reader.nextInt();
Scanner reader1 = new Scanner(System.in);
System.out.println("Enter the scend number");
// get user input for b
b = reader1.nextInt();
switch (input){
case 1: System.out.println(a + " + " + b + " = " + add(a, b));
break;
case 2: System.out.println(a + " - " + b + " = " + subtract(a, b));
break;
case 3: System.out.println(a + " / " + b + " = " + divide(a, b));
break;
case 4: System.out.println(a + " * " + b + " = " + multiply(a, b));
break;
default: System.out.println("your input is invalid!");
break;
}
}
static int add(int lhs, int rhs) { return lhs + rhs; }
static int subtract(int lhs, int rhs) { return lhs - rhs; }
static int divide(int lhs, int rhs) { return lhs / rhs; }
static int multiply(int lhs, int rhs) { return lhs * rhs; }
}

Scanner input = new Scanner(System.in);
String inputval = input.next();

Scanner input=new Scanner(System.in);
int integer=input.nextInt();
String string=input.next();
long longInteger=input.nextLong();

Just one extra detail. If you don't want to risk a memory/resource leak, you should close the scanner stream when you are finished:
myScanner.close();
Note that java 1.7 and later catch this as a compile warning (don't ask how I know that :-)

Here is a more developed version of the accepted answer that addresses two common needs:
Collecting user input repeatedly until an exit value has been entered
Dealing with invalid input values (non-integers in this example)
Code
package inputTest;
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");
boolean done = false;
while (!done) {
System.out.print("Enter an integer: ");
try {
int n = reader.nextInt();
if (n == 0) {
done = true;
}
else {
// do something with the input
System.out.println("\tThe number entered was: " + n);
}
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.
}
}
System.out.println("Exiting...");
reader.close();
}
}
Example
Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...
Note that without nextLine(), the bad input will trigger the same exception repeatedly in an infinite loop. You might want to use next() instead depending on the circumstance, but know that input like this has spaces will generate multiple exceptions.

import java.util.Scanner;
class Daytwo{
public static void main(String[] args){
System.out.println("HelloWorld");
Scanner reader = new Scanner(System.in);
System.out.println("Enter the number ");
int n = reader.nextInt();
System.out.println("You entered " + n);
}
}

Add throws IOException beside main(), then
DataInputStream input = new DataInputStream(System.in);
System.out.print("Enter your name");
String name = input.readLine();

It is very simple to get input in java, all you have to do is:
import java.util.Scanner;
class GetInputFromUser
{
public static void main(String args[])
{
int a;
float b;
String s;
Scanner in = new Scanner(System.in);
System.out.println("Enter a string");
s = in.nextLine();
System.out.println("You entered string " + s);
System.out.println("Enter an integer");
a = in.nextInt();
System.out.println("You entered integer " + a);
System.out.println("Enter a float");
b = in.nextFloat();
System.out.println("You entered float " + b);
}
}

import java.util.Scanner;
public class Myapplication{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int a;
System.out.println("enter:");
a = in.nextInt();
System.out.println("Number is= " + a);
}
}

You can get user input like this using a BufferedReader:
InputStreamReader inp = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(inp);
// you will need to import these things.
This is how you apply them
String name = br.readline();
So when the user types in his name into the console, "String name" will store that information.
If it is a number you want to store, the code will look like this:
int x = Integer.parseInt(br.readLine());
Hop this helps!

Can be something like this...
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
System.out.println("Enter a number: ");
int i = reader.nextInt();
for (int j = 0; j < i; j++)
System.out.println("I love java");
}

You can get the user input using Scanner. You can use the proper input validation using proper methods for different data types like next() for String or nextInt() for Integer.
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
//reads the input until it reaches the space
System.out.println("Enter a string: ");
String str = scanner.next();
System.out.println("str = " + str);
//reads until the end of line
String aLine = scanner.nextLine();
//reads the integer
System.out.println("Enter an integer num: ");
int num = scanner.nextInt();
System.out.println("num = " + num);
//reads the double value
System.out.println("Enter a double: ");
double aDouble = scanner.nextDouble();
System.out.println("double = " + aDouble);
//reads the float value, long value, boolean value, byte and short
double aFloat = scanner.nextFloat();
long aLong = scanner.nextLong();
boolean aBoolean = scanner.nextBoolean();
byte aByte = scanner.nextByte();
short aShort = scanner.nextShort();
scanner.close();

import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Welcome to the best program in the world! ");
while (true) {
System.out.print("Enter a query: ");
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
if (s.equals("q")) {
System.out.println("The program is ending now ....");
break;
} else {
System.out.println("The program is running...");
}
}
}
}

This is a simple code that uses the System.in.read() function. This code just writes out whatever was typed. You can get rid of the while loop if you just want to take input once, and you could store answers in a character array if you so choose.
package main;
import java.io.IOException;
public class Root
{
public static void main(String[] args)
{
new Root();
}
public Root()
{
while(true)
{
try
{
for(int y = 0; y < System.in.available(); ++y)
{
System.out.print((char)System.in.read());
}
}
catch(IOException ex)
{
ex.printStackTrace(System.out);
break;
}
}
}
}

I like the following:
public String readLine(String tPromptString) {
byte[] tBuffer = new byte[256];
int tPos = 0;
System.out.print(tPromptString);
while(true) {
byte tNextByte = readByte();
if(tNextByte == 10) {
return new String(tBuffer, 0, tPos);
}
if(tNextByte != 13) {
tBuffer[tPos] = tNextByte;
++tPos;
}
}
}
and for example, I would do:
String name = this.readLine("What is your name?")

Keyboard entry using Scanner is possible, as others have posted. But in these highly graphic times it is pointless making a calculator without a graphical user interface (GUI).
In modern Java this means using a JavaFX drag-and-drop tool like Scene Builder to lay out a GUI that resembles a calculator's console.
Note that using Scene Builder is intuitively easy and demands no additional Java skill for its event handlers that what you already may have.
For user input, you should have a wide TextField at the top of the GUI console.
This is where the user enters the numbers that they want to perform functions on.
Below the TextField, you would have an array of function buttons doing basic (i.e. add/subtract/multiply/divide and memory/recall/clear) functions.
Once the GUI is lain out, you can then add the 'controller' references that link each button function to its Java implementation, e.g a call to method in your project's controller class.
This video is a bit old but still shows how easy Scene Builder is to use.

The most simple way to get user input would be to use Scanner. Here's an example of how it's supposed to be used:
import java.util.Scanner;
public class main {
public static void main(String[]args) {
Scanner sc=new Scanner(System.in);
int a;
String b;
System.out.println("Type an integer here: ");
a=sc.nextInt();
System.out.println("Type anything here:");
b=sc.nextLine();
The line of code import java.util.Scanner; tells the program that the programmer will be using user inputs in their code. Like it says, it imports the scanner utility. Scanner sc=new Scanner(System.in); tells the program to start the user inputs. After you do that, you must make a string or integer without a value, then put those in the line a=sc.nextInt(); or a=sc.nextLine();. This gives the variables the value of the user inputs. Then you can use it in your code. Hope this helps.

Using JOptionPane you can achieve it.
Int a =JOptionPane.showInputDialog(null,"Enter number:");

import java.util.Scanner;
public class userinput {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Name : ");
String name = input.next();
System.out.print("Last Name : ");
String lname = input.next();
System.out.print("Age : ");
byte age = input.nextByte();
System.out.println(" " );
System.out.println(" " );
System.out.println("Firt Name: " + name);
System.out.println("Last Name: " + lname);
System.out.println(" Age: " + age);
}
}

class ex1 {
public static void main(String args[]){
int a, b, c;
a = Integer.parseInt(args[0]);
b = Integer.parseInt(args[1]);
c = a + b;
System.out.println("c = " + c);
}
}
// Output
javac ex1.java
java ex1 10 20
c = 30

How to read integer value from the standard input in Java

What class can I use for reading an integer variable in Java?

You can use java.util.Scanner (API):
import java.util.Scanner;
//...
Scanner in = new Scanner(System.in);
int num = in.nextInt();
It can also tokenize input with regular expression, etc. The API has examples and there are many others in this site (e.g. How do I keep a scanner from throwing exceptions when the wrong type is entered?).

If you are using Java 6, you can use the following oneliner to read an integer from console:
int n = Integer.parseInt(System.console().readLine());

Here I am providing 2 examples to read integer value from the standard input
Example 1
import java.util.Scanner;
public class Maxof2
{
public static void main(String args[])
{
//taking value as command line argument.
Scanner in = new Scanner(System.in);
System.out.printf("Enter i Value: ");
int i = in.nextInt();
System.out.printf("Enter j Value: ");
int j = in.nextInt();
if(i > j)
System.out.println(i+"i is greater than "+j);
else
System.out.println(j+" is greater than "+i);
}
}
Example 2
public class ReadandWritewhateveryoutype
{
public static void main(String args[]) throws java.lang.Exception
{
System.out.printf("This Program is used to Read and Write what ever you type \nType quit to Exit at any Moment\n\n");
java.io.BufferedReader r = new java.io.BufferedReader (new java.io.InputStreamReader (System.in));
String hi;
while (!(hi=r.readLine()).startsWith("quit"))System.out.printf("\nYou have typed: %s \n",hi);
}
}
I prefer the First Example, it's easy and quite understandable.
You can compile and run the JAVA programs online at this website: http://ideone.com

Check this one:
public static void main(String[] args) {
String input = null;
int number = 0;
try {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
input = bufferedReader.readLine();
number = Integer.parseInt(input);
} catch (NumberFormatException ex) {
System.out.println("Not a number !");
} catch (IOException e) {
e.printStackTrace();
}
}

Second answer above is the most simple one.
int n = Integer.parseInt(System.console().readLine());
The question is "How to read from standard input".
A console is a device typically associated to the keyboard and display from which a program is launched.
You may wish to test if no Java console device is available, e.g. Java VM not started from a command line or the standard input and output streams are redirected.
Console cons;
if ((cons = System.console()) == null) {
System.err.println("Unable to obtain console");
...
}
Using console is a simple way to input numbers. Combined with parseInt()/Double() etc.
s = cons.readLine("Enter a int: ");
int i = Integer.parseInt(s);
s = cons.readLine("Enter a double: ");
double d = Double.parseDouble(s);

check this one:
import java.io.*;
public class UserInputInteger
{
public static void main(String args[])throws IOException
{
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int number;
System.out.println("Enter the number");
number = Integer.parseInt(in.readLine());
}
}

This causes headaches so I updated a solution that will run using the most common hardware and software tools available to users in December 2014. Please note that the JDK/SDK/JRE/Netbeans and their subsequent classes, template libraries compilers, editors and debuggerz are free.
This program was tested with Java v8 u25. It was written and built using
Netbeans IDE 8.0.2, JDK 1.8, OS is win8.1 (apologies) and browser is Chrome (double-apologies)
- meant to assist UNIX-cmd-line OG's deal with modern GUI-Web-based IDEs
at ZERO COST - because information (and IDEs) should always be free.
By Tapper7. For Everyone.
code block:
package modchk; //Netbeans requirement.
import java.util.Scanner;
//import java.io.*; is not needed Netbeans automatically includes it.
public class Modchk {
public static void main(String[] args){
int input1;
int input2;
//Explicity define the purpose of the .exe to user:
System.out.println("Modchk by Tapper7. Tests IOStream and basic bool modulo fxn.\n"
+ "Commented and coded for C/C++ programmers new to Java\n");
//create an object that reads integers:
Scanner Cin = new Scanner(System.in);
//the following will throw() if you don't do you what it tells you or if
//int entered == ArrayIndex-out-of-bounds for your system. +-~2.1e9
System.out.println("Enter an integer wiseguy: ");
input1 = Cin.nextInt(); //this command emulates "cin >> input1;"
//I test like Ernie Banks played hardball: "Let's play two!"
System.out.println("Enter another integer...anyday now: ");
input2 = Cin.nextInt();
//debug the scanner and istream:
System.out.println("the 1st N entered by the user was " + input1);
System.out.println("the 2nd N entered by the user was " + input2);
//"do maths" on vars to make sure they are of use to me:
System.out.println("modchk for " + input1);
if(2 % input1 == 0){
System.out.print(input1 + " is even\n"); //<---same output effect as *.println
}else{
System.out.println(input1 + " is odd");
}//endif input1
//one mo' 'gain (as in istream dbg chk above)
System.out.println("modchk for " + input2);
if(2 % input2 == 0){
System.out.print(input2 + " is even\n");
}else{
System.out.println(input2 + " is odd");
}//endif input2
}//end main
}//end Modchk