Java, Exception works properly but the user does not get the chance to input again and in console the programs doesn't stop

Question

I want the user to put only integers and if he enters anything else he should get my error and should be asked to input again. But this code is not working as I want my program to be. Please help me for this and other suggestion are also welcome.

//this is a separate main file 

import java.util.*;

public class SolMul {

    public static void main(String[] args) {


        int g = 1;

        Scanner input = new Scanner(System.in);

        do{

        try{

        System.out.println("Please enter the number you want the table for: \n");

        int z = input.nextInt();

        System.out.println("Upto what number you want the table:\n");
        int y = input.nextInt();


        Solve multiplyObj = new Solve();
        multiplyObj.multiply(z, y);




        g = 2;
        }catch(InputMismatchException  e){
            System.out.println("Error");
        }


        }

        while(g==1);


        input.close();
    }


}

//this is a separate class file in a new window


public class Solve {


    public  void multiply(int number,int upto){


        for (int x = 1; x <= upto; x ++){
            System.out.printf("%d X %d = %d  \n",number,x,number*x );
        }

    }
}

What is it doing (wrong) ? And what exactly do you want it to do differently? — Stultuske, Sep 09 '14 at 11:20
Well, have you stepped through the code in your IDE's debugger? — OldProgrammer, Sep 09 '14 at 11:23
in case we input a decimal or a any alphabet, instead of showing error the console keeps on doing this contionously "error "Please enter the number you want the table for: \n" — Harpreet Singh Saini, Sep 09 '14 at 11:25
If you used a debugger, you could find out yourself. http://stackoverflow.com/questions/25385173/what-is-a-debugger-and-how-can-it-help-me-diagnose-problems — Raedwald, Sep 09 '14 at 12:08

Dr. Debasish Jana · Answer 1 · 2014-09-09T12:08:20.913

0

In addition to

    catch(InputMismatchException  e){
                System.out.println("Input Mismatch Error");
}

Add another after that:

    catch(Exception  e){
                System.out.println("Other Error");
}

Instead of input.nextLine(), use input.next()

Then check again

edited Sep 09 '14 at 12:08

answered Sep 09 '14 at 11:26

Dr. Debasish Jana

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it didnt work after that to, its shows the error is caught bt the programs keeps on continuing and then program shuts down – Harpreet Singh Saini Sep 09 '14 at 11:31

score 0 · Accepted Answer · answered Sep 09 '14 at 11:40

First,

input.nextInt()

doesn't capture newline. So you need to replace that with input.nextLine().

            int z = Integer.parseInt(input.nextLine());
            int y = Integer.parseInt(input.nextLine());

You can add an Exception catch block to handle any parsing errors. Making those changes to your code.

public static void main(String[] args) {
    int g = 1;
    Scanner input = new Scanner(System.in);
    do {
        try {
            System.out.println("Please enter the number you want the table for: \n");
            int z = Integer.parseInt(input.nextLine());
            System.out.println("Upto what number you want the table:\n");
            int y = Integer.parseInt(input.nextLine());
            Solve multiplyObj = new Solve();
            multiplyObj.multiply(z, y);
            g = 2;
        } catch (InputMismatchException e) {
            System.out.println("Error -- input again");
        }catch (Exception e) {
            System.out.println("Error -- input again");
        }           
    }while (g == 1);
    input.close();
}

Java, Exception works properly but the user does not get the chance to input again and in console the programs doesn't stop

2 Answers2