I am trying to check if a user entered a number and if not, then it uses a default number like 10. How do i check if the user just presses enter and doesn't enter anything in Java
input = scanner.nextInt();
pseudo code:
if(input == user just presses enter without entering anything){
input = 10;
}
else just proceed with input = what user entered
//scanner is a Scanner
int i; // declare it before the block
try {
i = scanner.nextInt();
}
catch (InputMismatchException ime) {
i = 10;
}
// i is some integer from the user, or 10
First things first, geeeeeez guys, when the OP says something like
"I don't want an exception, i want i = 10 if nothing is entered, so what do i do"
That should clue you in that he probably doesn't know too much about exceptions (maybe even java) and might need a simple answer. And if that's not possible, explain to him the difficult ones.
Alright, here's the plain and simple way to do it
String check;
int input = 10;
check = scanner.nextLine/*Int*/();
if(check.equals(""))
{
//do nothing since input already equals 10
}
else
{
input = Integer.parseInt(check);
}
Let me explain what this code is doing. You were originally using nextInt() to get your number for input, correct? The problem is, nextInt() only responds if the user actually inputs something, not if they press enter. In order to check for enter, we used a method that actually responds when the user presses enter and used that to ensure that our code does what we wanted to. One thing I recommend using is an API, Java has one.
Here's the link for the API HERE
And here's the link for the actual method I used HERE. You can find descriptions and instructions on many methods you'll run into on this API.
Now, back to my answer, that's the easy way to do it. Problem is, this code isn't necessarily safe. It'll throw exceptions if something goes wrong, or if someone is trying to hack into your system. For example, if you were to enter a letter instead of pressing enter or entering a number, it would throw an exception. What you've been seeing in the other answers is what we call exception handling, that's how we make sure exceptions don't happen. If you want an answer that'll catch most of these exceptions, you need to make sure your code catches them, or avoids them all together (I'm simplifying things immensely). The above answer is working code, but isn't safe code, you wouldn't ever use something like this all by itself in real life.
Here is something that might be considered safe code. And no exceptions to keep it simple! ;)
import java.util.Scanner;
public class SOQ15
{
public Scanner scanner;
public SOQ15()
{
scanner = new Scanner(System.in);
int input = 10;
boolean isAnInt = true;
String check;
check = scanner.nextLine/*Int*/();
if(check.equals(""))
{
//do nothing since input already equals 10
}
for(int i = 0; i < check.length(); i++)
{
if(check.charAt(i) >= '0' && check.charAt(i) <= '9' && check.length() < 9)
{
//This is if a number was entered and the user didn't just press enter
}
else
{
isAnInt = false;
}
}
if(isAnInt)
{
input = Integer.parseInt(check);
System.out.println("Here's the number - " + input);
}
}
public static void main(String[] args)
{
SOQ15 soq = new SOQ15();
}
}
I don't have time to go into all the details right now, but ask and I'll gladly respond when I get the time! :)
Well if you are using scanner, given the details provided, you can try:
Scanner in = new Scanner(System.in);
if in.hasNextInt(){ //Without this, the next line would throw an input mismatch exception if given a non-integer
int i = in.nextInt(); //Takes in the next integer
}
You said you wanted a default of 10 otherwise so:
else {
int i = 10;
}
Related
I want to add an integer to a list based on user input. The user has to type all the integers he/she wishes then press enter. if they finish inputting integer, they are supposed to press the "enter" button without typing anything.
I have made my code, but there are several mistakes
the exception keeps popping up because every time say for example I enter integer 10, then I finish. I press "enter" with nothing. this raises the exception. how do I tackle this problem?
and another thing, how do I make the program so that if the user puts invalid input, instead of crashing or breaking. It asks the user again to prompt the correct input.
this is what I have done
package basic.functions;
import java.util.*;
import java.text.DecimalFormat;
public class Percent {
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
reader.useDelimiter(System.getProperty("line.separator"));
List<Integer> list = new ArrayList<>();
System.out.println("Enter Integer: ");
while (true) {
try {
int n = reader.nextInt();
list.add(Integer.valueOf(n));
} catch (InputMismatchException exception) {
System.out.println("Not an integer, please try again");
break;
}
}
reader.close();
}
}
output
Enter Integer:
10
Not an integer, please try again
[10]
I'd suggest you utilise Scanner#hasNextInt to identify whether an integer has been entered or not. As for when the "user presses enter without typing anything", we can simply use the String#isEmpty method.
while (true) {
if(reader.hasNextInt()) list.add(reader.nextInt());
else if(reader.hasNext() && reader.next().isEmpty()) break;
else System.out.println("please enter an integer value");
}
note - in this case, you don't need to catch InputMismatchException because it won't be thrown.
while (true) is generally a bad sign, if you ever have that in your code you are almost certainly wrong.
What you probably want is something like this:
String input;
do {
input = reader.next();
// Parse the input to an integer using Integer.valueOf()
// Add it to the list if it succeeds
// You will need your try/catch etc here
while (!input.isEmpty());
Here the loop is checking the exit condition and running until it meets it. Your processing is still done inside the loop as normal but the program flow is a lot cleaner.
I'm a newbie in Java. I started these days and I'm practicing the catch and try exception. I have this code below which solve an operation between to numbers and I'd like to know what can I do, if for example I want that the user, once he makes an operation and get his result, that this has the possibility to make another operation. something like comes up a question asking if he wants to realize another problem and the code run again from the beginning.
package justpractice;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner Operation = new Scanner(System.in);
int x=1;
while(x==1){
try{
System.out.println("Insert numerator");
int n1 = Operation.nextInt();
System.out.println("Insert denominator");
int n2=Operation.nextInt();
double division = n1/n2;
System.out.println(division);
x=2;
}
catch(Exception e){
System.out.println("Insert a valid value");
}
}
}
}
You can do, for example, adding an if statement with a
System.out.println("Do you want to recalculate ? (1/0 Yes/No)");
Operation.nextInt();
then if the input is 1, keep x = 1, else do x = 2.
Try this code amendment;
package justpractice;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner Operation = new Scanner(System.in);
//better practice
while(true){
try{
System.out.println("Insert numerator");
int n1 = Operation.nextInt();
System.out.println("Insert denominator");
int n2=Operation.nextInt();
double division = n1/n2;
System.out.println(division);
System.out.println("Continue? (y/n)");
String response = Operation.nextLine();
if (response.equals("n")){
break;
}
}
catch(Exception e){
System.out.println("Insert a valid value");
}
}
}
}
To allow your user to calculate division again, you could use a do-while loop. With this loop you can execute the code inside once, ask the user if they would like to calculate again, and repeat the code if they do.
An outline of your code with the loop would something like this:
...
boolean again = true;
do { //Main loop
try {
... //Your division code
... //Put code to ask the user if they want to calculate again here
} catch(Exception e) {
...
}
} while(again == true); //Repeat if the user wants to continue
To get input on if the user wants to calculate again, I recommend using another do-while loop with your Scanner. This loop will allow you to repeat the code when the answer is invalid. In this case, when it's not "y" or "n".
String input;
do {
System.out.println("Would you like to continue? (y/n)");
input = operation.next();
} while(!input.equalsIgnoreCase("y") && !input.equalsIgnoreCase("n"));
After you have got the user's input, you still need to terminate the loop if they said "n". To do this you could use an if statement.
if(input.equalsIgnoreCase("n")) {
again = false; //Terminate the main loop
operation.close(); //Prevent any resource leaks with your Scanner
}
There is no need to check if the user input "y" as again is set to true by default.
Side Note: Variables should always be camelCase. Look at the Java Naming Conventions to learn more about naming things in Java.
EDIT:
The reason the console is repeatedly logging that you entered a non-number even though you entered it once, I'm not exactly sure. I think it's because the call to nextInt() never finishes because of the InputMismatchException being thrown, causing the next call to nextInt() (After the do-while repeats) to think that the letter/symbol you just entered is the one you want to process, repeating the exception over and over again.
To solve this, add this line into your catch block:
if(operation.hasNext()) operation.next();
This will call next() and complete the process of marking the letter/symbol you just entered as already processed, then repeat the do-while loop as normal.
In my console application, I give the user the option to type "exit" at any point to return to the main menu. As the user enters data, i prompt him/her through the console for various things, and collect the data using scanner.
My question is how can I check to see if the user entered "exit" after each prompt (as opposed to the requested information) without having to use the same if statement after each step?
As I see it, any kind of while or for loops are insufficient because they only check the condition at the beginning, when I need to check the condition between inputs, and I need each input/prompt to execute only once per iteration.
The key here is that each prompt/code executed between the checks is DIFFERENT, so loops won't work.
Here is some example code:
String first;
String second;
Scanner input = new Scanner(System.in);
//prompt user for input
first = input.nextline();
if(first.equals("exit")){
//return to start menu
input.close();
return;
}
//prompt user for DIFFERENT input
second = input.nextline()
if(second.equals("exit")){
//return to start menu
input.close();
return;
}
If I understand you correctly, I recommend a do-while loop. It will first take text the first time, perform an action, and if it is exit it will break the loop, otherwise it will repeat.
do{
text = input.nextLine();
//whatever code you want here to perform with input
}while(!(text.equals("exit"));
Write a method...
public boolean isExit(String value) {
return value.equals("exit");
}
You can then check this method each time...
String value = input.nextLine();
if (!isExit(value) {
// Handle the normal text
} else {
// Handle the exit operations...
}
You could put additional code in the check, but I would prefer to have an additional method that handles the exit operation...for example...
String value = input.nextLine();
if (!isExit(value) {
// Handle the normal text
} else {
doExit();
}
Take a look at Defining Methods for more details...
Updated
Focus on the idea that a method should do a single job and have no side effects...
Having said that, I would setup my code in such away that if the user enters exit at the prompt, the method can exit of it's own accord, without the need for return; statement...
For example...
public int selectYourMeal() {
// Prompt...
int option = -1;
String value = input.nextLine();
if (!isExit(value) {
// Handle the normal text
} else {
option = EXIT_OPTION;
}
return option;
}
Where EXIT_OPTION is a special value, which the caller and identify and deal with as it sees fit.
I'm also old school, in that I was taught that a method should have one entry and one exit point, you really want to avoid having multiple exit points within your methods, as it becomes very difficult to follow the logic...
I would suggest you use an List<String> of words. You can return it. Also, it's a really bad idea to close a Scanner wrapping System.in, once you do you cannot re-open it.
List<String> words = new ArrayList<>();
Scanner input = new Scanner(System.in);
for (int i = 0; input.hasNextLine(); i++) {
String line = input.nextLine();
words.add(line);
if (line.equalsIgnoreCase("exit")) {
break;
}
}
System.out.println(words);
return words;
this i my first attempt at asking a question so hopefully it shows correctly. Basically what I need the program to do is to ask the user for a preset account number and password and only allow them 3 attempts. I then want to call up another method when both requirements are met so i can continue with the program. The first problem i have is that when i enter the correct password its is still showing as incorrect and i don't know why, then i would like to know if i have call the method within the if statement correctly. Thanks.
import java.util.Scanner;
public class Part4 {
public static void main(String[] args)
{
String password = "password", passwordattempt = null;
int accnum = 123456789, acctry = 0, tries = 0;
Scanner input = new Scanner (System.in);
while (acctry != accnum){
System.out.println("\nPlease enter your account number");
acctry = input.nextInt();
if (acctry != accnum)
System.out.print("That number is incorrect. Please try again.");
else
if (acctry == accnum)
{
while (tries < 3)
{
System.out.println("\nPlease enter password");
passwordattempt = input.next();
if (passwordattempt != password){
System.out.print("That password is incorrect");
tries++;
}
else
if (passwordattempt == password){
System.out.print("That is correct");
AccountDetails.Details(args);
}
}
System.out.print("\nYou have exceeded the ammount of tries");
}
}
}
public static class AccountDetails {
private static void Details(String[] args){
System.out.print("it works");
}
}
}
two problems.
1: You're executing your while loop regardless of if it is successful or not.
.
while(tries < 3)
should be
while(tries < 3 && !successfulPassword)
You'll need to add the successfulPassword variable, so that you don't get it right the first time and yet continue to have to enter passwords.
2: Your comparison of strings is grossly, umm, well, wrong. There's two things that catch my eye. The first is you can't use == and != and get the results you expect. You must use .equals(). Secondly, you don't need to repeat the opposite clause like you do with a human. For example, I tell my daughter "If you eat your supper, then you may have cookies. Else, if you do not eat your supper, then you may not have cookies." To a computer, you don't need that last "if you do not eat your supper". It's guaranteed to be true (since you're in the else block anyway) and it just clutters it up. So that just becomes
.
if(passwordAttempt.equals(password) {
successfulPassword = true;
} else {
tries++;
}
In the Java language, Strings are objects, and thus comparing them using '==' is testing by reference, and not by equality.
I believe what you are looking for is
if (passwordattempt.equals(password)) {
Check here for more information:
http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#equals(java.lang.Object)
For an assignment I have due, my group and I were asked to code an educational/interactive game, and we decided on a basic maths one.
To get the users answers, we decided to use Java Scanner and put this line of code at the top of all the code we have;
java.util.Scanner
One of the loops that use this is the page with the questions on it, the loop looking something like this;
scoreCount = 0;
for (questions = 0; questions < 5;) {
//get the user's answer
userAnswer[questions] = input.nextInt();
//text box for users answer
if (userAnswer[questions] == compAnswer) {
//put tick next to answer
//add one to score
scoreCount = scoreCount + 1;
} else if (userAnswer[questions] != compAnswer) {
//put cross next to answer
}
//go to next question
questions++ ;
}
I'm working through all the errors that were thrown up and every time i don't have java.util.Scanner commented out Processing throws us the errors unexpected token: and then either class or void, which i don't get, but when java.util.Scanner is commented out, the classes and voids all work and the .input.nextInt() isn't recognised.
I am new to Java programming and Processing, any help at all would be greatly appreciated
EDIT
i think this is the link which lets you see my code, it's called Test;
https://github.com/MeganSime/Week8DataVis
you have to check if scanner has next int (token)
Scanner input = new Scanner(System.in);
.
.
if(input.hasNextInt()) { // or hasNext()
userAnswer[questions] = input.nextInt();
}
You're probably inserting a non int value where the scanner expects that. You should do something like that:
if(input.hasNextInt()) {
userAnswer[questions] = input.nextInt();
} else {
scan.next(); //consume any non-int value like ":"
}