We Keep Coding

When user presses "enter" key or space enter then error message pops up

I would like to print an error message when the user presses enter or space enter instead of a string. I have tried isEquals("") and isEmpty() but haven't found anything that works yet.
Here's my code:
import java.util.Scanner;
public class check{
public static void main(String args[]){
System.out.println("Enter a number: ");
Scanner keyboard = new Scanner(System.in);
String input = keyboard.next();
if(input.equals("")){
System.out.println("Empty");
} else {
System.out.println("number inputed");
}
}
}

One way to do this, change keyboard.next() to keyboard.nextLine(), use trim() to remove unnecessary spaces, check with isEmpty().
String input = keyboard.nextLine().trim();
if (input.isEmpty()) {
// error message
} else {
// good to go
}

import java.util.Scanner;
public class check{
public static void main(String args[]){
System.out.println("Enter a number: ");
Scanner keyboard = new Scanner(System.in);
String input = keyboard.nextLine();
if(input.trim().equals("")){
System.out.println("Empty");
} else {
System.out.println("number inputed");
}
}
}

Strangely, I don't get an error when running your code. However, I noticed that your code simply doesn't react to an empty input (just pressing enter). If you want to check for that, you can use keyboard.nextLine().
Judging by the rest of your code, it seems like you want the user to input only a number. An easy way to check if the user entered an integer if you're using Scanner is keyboard.hasNextInt().
Meaning you can do something like this:
if(keyboard.hasNextInt()) {
int yourNumber = keyboard.nextInt();
System.out.println("Your number is: " + your Number);
}
else {
System.out.println("Please enter a valid integer");
}

To check whether the string input is empty, you can use the String.isEmpty() method. Look below:
String input = keyboard.nextLine();
if(!input.isEmpty()) {
//the input is not empty!
}
else {
//the input is empty!
}
Note, however, that since you want to receive numbers as inputs you should not retrieve them as strings. Below is an example where the program retrieves a double from the user. Scanner provides many methods to validate the user's input. In this case, I'm using hasNextDouble() to check whether the input is a number.
Scanner scanner = new Scanner(System.in);
System.out.println("Enter a number:");
while(!scanner.hasNextDouble()) {
System.out.println("That's not a number!");
scanner.next();
}
double numberInput = scanner.nextDouble();
System.out.println("The entered number was " + numberInput);

I made a sample program similar to yours and used nextLine() instead of next(). When user enters space and clicks enter he will print "space" else "a number".

StringOutOfBoundsException 1 Java

This is what the error says:
java.lang.StringIndexOutOfBoundsException: String index out of range: 1
at java.lang.String.substring(String.java:1963)
at FracCalc.produceAnswer(FracCalc.java:34)
at FracCalc.main(FracCalc.java:16)
And Here is my existing code:
import java.util.*;
public class FracCalc {
public static void main(String[] args)
{
// TODO: Read the input from the user and call produceAnswer with an equation
Scanner input = new Scanner(System.in);
System.out.println("Do you want to calculate something? If no, type in 'quit'. If yes, simply type in 'yes'. ");
String Continue = input.next();
if(Continue.equals("quit")){
System.out.println("Have a nice day! ");
}else if(Continue.equals("yes")){
System.out.println("Please input a fraction, there must be exactly one space between the operator and the operand. ");
String readInput = input.nextLine();
System.out.println(produceAnswer(readInput));
}
}
// ** IMPORTANT ** DO NOT DELETE THIS FUNCTION. This function will be used to test your code
// This function takes a String 'input' and produces the result
//
// input is a fraction string that needs to be evaluated. For your program, this will be the user input.
// e.g. input ==> "1/2 + 3/4"
//
// The function should return the result of the fraction after it has been calculated
// e.g. return ==> "1_1/4"
public static String produceAnswer(String input)
{
// TODO: Implement this function to produce the solution to the input
int space = input.indexOf(" ");
int nextspace = space + 2;
String operand = input.substring(0, space + 1);
String operator = input.substring(space + 1, nextspace);
String operand2 = input.substring(nextspace + 1, input.length());
return operand2;
}
// TODO: Fill in the space below with any helper methods that you think you will need
}
I'm trying to segregate fraction operators by determining the indexes of spaces and then going from there. Unfortunately i keep getting this error. Can someone please help! Thank You!

You have to use input.nextLine() since you want to read the entire line and not just characters. Try this in your main method:
Scanner input = new Scanner(System.in);
System.out.println("Do you want to calculate something? If no, type in 'quit'. If yes, simply type in 'yes'. ");
String Continue = input.nextLine();
if (Continue.equals("quit")) {
System.out.println("Have a nice day! ");
} else if (Continue.equals("yes")) {
System.out.println(
"Please input a fraction, there must be exactly one space between the operator and the operand. ");
String readInput = input.nextLine();
System.out.println(produceAnswer(readInput));
}

How do I use java.util.Scanner with integers

I just started learning java and I want to make a simple program where it requests the persons
name, outputs the name then asks for the thier favorite number. It will then compare their number
to the number 6 and will output something depending on if the number is larger or smaller than 6.
I am getting a "String to int convert" error in Netbeans which has to do with the scanner.
Hopefully I am asking this correctly but how can I make the scanner pick-up integers?
Thanks
package javaapplication2;
import java.util.Scanner;
import java.lang.String;
public class JavaApplication2 {
public static void main(String[] args) {
// Creating an instance of the scanner class.
// Gets name and numbers.
Scanner getName = new Scanner(System.in);
Scanner getNumber = new Scanner(System.in);
//Holds name and number
String userName;
int userNumber;
// Asks for the users name.
// Holds name in userName.
System.out.println("What is your name?");
userName = getName.nextLine();
//Reponds with the users name.
System.out.println("Hello" + userName + "!");
//Asks for favorite number.
// Holds number in userNumber.
System.out.println("What is your favorite number?");
userNumber = getNumber.nextLine();
// Checks if users number is larger than 6.
if (userNumber > 6) {
// Stuff goes here.
}
}
}

You should use only one Scanner for one input stream:
Scanner in = new Scanner(System.in);
And after that you should use it's methods to get integers:
String name = in.nextLine();
int number = in.nextInt();
To be sure, you should read the documentation for Scanner:
Scanner
Scanner::nextLine
Scanner::nextInt

This might help: Javadoc page for Scanner.

Check if string taken with Scanner.next() contains number

I wanted to take a String using Scanner.next() and see if it contains a number or not. I used regex to check if a string contains anything but a number. The regex works correctly if the string is hard coded, but not when taken from keyboard. I expected input of 5 to be detected as a number, but it is not. Please tell me why. My code:
import java.util.Scanner;
public class Error {
public static void main(String[] args) {
Scanner inp = new Scanner(System.in);
int num = 0;
String input = "";
boolean isStringNumber = true;
System.out.println("\nPlease enter a number only...");
input = inp.nextLine();
isStringNumber = input.contains("[^0-9]+");
if (isStringNumber == false) {
System.out.println("\nYou entered a non number " + input);
}
}
}

contains uses a String literal as its argument. Use matches instead
isStringNumber = input.matches("[0-9]+");
or simply
isStringNumber = input.matches("\\d+");
BTW: Scanner has a nextInt method for accepting integer values

instead isStringNumber = input.contains("[^0-9]+");
try isStringNumber = input.matches("[0-9]+");

Restrict string input from user to alphabetic and numerical values [duplicate]

This question already has an answer here:
How to use java.util.Scanner to correctly read user input from System.in and act on it?
(1 answer)
Closed 5 years ago.
Basically, my situation requires me to check to see if the String that is defined by user input from the keyboard is only alphabetical characters in one case and only digits in another case. This is written in Java.
my current code:
switch (studentMenu) {
case 1: // Change all four fields
System.out.println("Please enter in a first name: ");
String firstNameIntermediate = scan.next();
firstName = firstNameIntermediate.substring(0,1).toUpperCase() + firstNameIntermediate.substring(1);
System.out.println("Please enter in a middle name");
middleName = scan.next();
System.out.println("Please enter in a last name");
lastName = scan.next();
System.out.println("Please enter in an eight digit student ID number");
changeID();
break;
case 2: // Change first name
System.out.println("Please enter in a first name: ");
firstName = scan.next();
break;
case 3: // Change middle name
System.out.println("Please enter in a middle name");
middleName = scan.next();
break;
case 4: // Change last name
System.out.println("Please enter in a last name");
lastName = scan.next();
case 5: // Change student ID:
changeID();
break;
case 6: // Exit to main menu
menuExit = true;
default:
System.out.println("Please enter a number from 1 to 6");
break;
}
}
}
public void changeID() {
studentID = scan.next();
}
I need to make sure the StudentID is only numerical and each of the name segments are alphabetical.

java.util.Scanner can already check if the next token is of a given pattern/type with the hasNextXXX methods.
Here's an example of using boolean hasNext(String pattern) to validate that the next token consists of only letters, using the regular expression [A-Za-z]+:
Scanner sc = new Scanner(System.in);
System.out.println("Please enter letters:");
while (!sc.hasNext("[A-Za-z]+")) {
System.out.println("Nope, that's not it!");
sc.next();
}
String word = sc.next();
System.out.println("Thank you! Got " + word);
Here's an example session:
Please enter letters:
&###$
Nope, that's not it!
123
Nope, that's not it!
james bond
Thank you! Got james
To validate that the next token is a number that you can convert to int, use hasNextInt() and then nextInt().
Related questions
Validating input using java.util.Scanner - has many examples!

It's probably easiest to do this with a regular expression. Here's some sample code:
import java.util.regex.*;
public class Test
{
public static void main(String[] args) throws Exception
{
System.out.println(isNumeric("123"));
System.out.println(isNumeric("abc"));
System.out.println(isNumeric("abc123"));
System.out.println(isAlpha("123"));
System.out.println(isAlpha("abc"));
System.out.println(isAlpha("abc123"));
}
private static final Pattern NUMBERS = Pattern.compile("\\d+");
private static final Pattern LETTERS = Pattern.compile("\\p{Alpha}+");
public static final boolean isNumeric(String text)
{
return NUMBERS.matcher(text).matches();
}
public static final boolean isAlpha(String text)
{
return LETTERS.matcher(text).matches();
}
}
You should probably write methods of "getAlphaInput" and "getNumericInput" which perform the appropriate loop of prompt/fetch/check until the input is correct. Or possibly just getInput(Pattern) to avoid writing similar code for different patterns.
You should also work out requirements around what counts as a "letter" - the above only does a-z and A-Z... if you need to cope with accents etc as well, you should look more closely at the Pattern docs and adapt appropriately.
Note that you can use a regex to validate things like the length of the string as well. They're very flexible.

Im not sure this is the best way to do, but you could use Character.isDigit() and Character.IsLiteral() mabybe like this:
for( char c : myString.toCharArray() ) {
if( !Character.isLiteral(c) ) {
//
}
}

try regexp: \d+ -- numerical, [A-Za-z]+ -- alphabetical

I don't think you can prevent the users from entering invalid values, but you have the option of validating the data you receive. I'm a fan of regular expressions. Real quick, something like this maybe (all values initialized to empty Strings):
while (!firstName.matches("^[a-zA-Z]+$")) {
System.out.println("Please enter in a first name");
firstName = scan.next();
}
...
while (!studentID.matches("^\\d{8}$")) {
System.out.println("Please enter in an eight digit student ID number");
changeID();
}
If you go this route, you might as well categorize the different cases you need to validate and create a few helper methods to deal with each.
"Regex" tends to seem overwhelming in the beginning, but learning it has great value and there's no shortage of tutorials for it.

That is the code
public class InputLetters {
String InputWords;
Scanner reader;
boolean [] TF;
boolean FT;
public InputLetters() {
FT=false;
while(!FT){
System.out.println("Enter that you want to: ");
reader = new Scanner(System.in);
InputWords = reader.nextLine();
Control(InputWords);
}
}
public void Control(String s){
String [] b = s.split(" ");
TF = new boolean[b.length];
for(int i =0;i<b.length;i++){
if(b[i].matches("^[a-zA-Z]+$")){
TF[i]=true;
}else
{
TF[i]=false;
}
}
for(int j=0;j<TF.length;j++){
if(!TF[j]){
FT=false;
System.out.println("Enter only English Characters!");
break;
}else{
FT=true;
}
}
}