How to parse int from String

Question

I have to keep inputting x and y coordinates, until the user inputs "stop". However, I don't understand how to parse the input from String to int, as whenever I do, I get back errors.

public class Demo2 {
    public static void main(String[] args) {

        Scanner kb = new Scanner(System.in);

        while (true) {
            System.out.println("Enter x:");
            String x = kb.nextLine();

            if (x.equals("stop")) {
                System.out.println("Stop");
                break;
            }

            System.out.println("Enter y:");
            String y = kb.nextLine();

            if (y.equals("stop")) {
                System.out.println("Stop"); }
                break;
            }
        }
    }
}

Integer.parseInt(x) - but this will throw an exception if the value is not a integer value — MadProgrammer
– MadProgrammer, Commented Sep 22, 2015 at 0:07
Title is inadequate to the content: should be "parse this string" — Jacek Cz
– Jacek Cz, Commented Sep 22, 2015 at 0:30

ashiquzzaman33 · Accepted Answer · 2015-09-22 00:21:00Z

1

To Parse integer from String you can use this code snippet.

    try{

     int    xx = Integer.parseInt(x);
     int    yy = Integer.parseInt(y);

        //Do whatever want
    }catch(NumberFormatException e){
        System.out.println("Error please input integer.");
    }

answered Sep 22, 2015 at 0:21

ashiquzzaman33

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Comments

Hips · Accepted Answer · 2015-09-22 00:28:23Z

1

Nice way to do this in my opinion is to always read the input as a string and then test if it can be converted to an integer.

import java.util.Scanner;

public class Demo2 {
    public static void main(String[] args) {
        Scanner kb = new Scanner(System.in);

        while (true) {
            String input;
            int x = 0;
            int y = 0;
            System.out.println("Enter x:");
            input = kb.nextLine();

            if (input.equalsIgnoreCase("STOP")) {
                System.out.println("Stop");
                break;
            }

            try {
                x = Integer.parseInt(input);
                System.out.println(x);
            } catch (NumberFormatException e) {
                System.out.println("No valid number");
            }
        }
    }
}

edited Sep 22, 2015 at 0:28

answered Sep 22, 2015 at 0:21

Hips

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1 Comment

codekid Over a year ago

Thanks, this helped alot

Tom · Accepted Answer · 2015-09-22 00:24:13Z

0

With

String variablename = Integer.toString(x);

edited Sep 22, 2015 at 0:24

Tom

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answered Sep 22, 2015 at 0:13

Mike Hurtado A

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3 Comments

TNT Over a year ago

This won't work; toString requires an int, not a String.

Mike Hurtado A Over a year ago

ok, int xx = Integer.parseInt(x); and String variablename = Integer.toString(xx); if(variablename.equals("stop")){...

TNT Over a year ago

Then why go through the trouble of converting the String to an int and back? Besides, assuming Integer.parseInt does not throw an exception, the resultant string will never not be a number, so variableName.equals("stop") will always return false.

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