INVALID INPUT AND LOOP

Question

T student Im wondering if the user input a Character not an integer.How can i Show the word INVALID to him and let him Type again.

EXAMPLE:

input a two number

a

INVALID try Again:1

2

Sum of two number=3

public static void main(String[] args) {
    int x = 0, y, z;

    System.out.println("Enter two integers to calculate their sum");
    Scanner in = new Scanner(System.in);

    try {
        x = in.nextInt();
    } catch (InputMismatchException e ) {
        System.out.print("INVALID");
    }

    y = in.nextInt();
    z = x + y;

    System.out.println("Sum of the integers = " + z);
}

you have to use a `while` loop and `break` out of it once the input is valid — Lino, Aug 08 '18 at 09:38
[Validating input using java.util.Scanner](https://stackoverflow.com/q/3059333) — Pshemo, Aug 08 '18 at 10:02

Veselin Davidov · Answer 1 · 2018-08-08T10:30:13.377

1

You can do for example:

            while(true) {
                try {
                    x = in.nextInt();
                    break;
                  } catch (InputMismatchException e ) {
                    System.out.print("INVALID try again:");
                    in.next(); //To wait for next value otherwise infinite loop
                  }
            }

Basically you need to add the input into a loop and keep looping until you get valid input. You can write it in different ways but that should be the idea.

The in.next() in the catch is required because nextInt() doesn't consume the new line character of the first input and this way we skip to that.

If I were you I would use in.nextLine() for each line of parameters and the manipulate the String that I get to check for valid input instead of waiting for exception.

edited Aug 08 '18 at 10:30

answered Aug 08 '18 at 09:38

Veselin Davidov

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just use `break` instead of introducing a variable – Lino Aug 08 '18 at 09:39
There is a problem – Hero Man Aug 08 '18 at 09:42
Non stop INVALID try again:INVALID try again:INVALID try again:INVALID try again:INVALID try again: – Hero Man Aug 08 '18 at 09:45
ok I eddited it ;) I didn't actually test it I just wanted to give the idea. get the version now. You need to add one in.next() in order to move to next input – Veselin Davidov Aug 08 '18 at 09:49
@Stultuske if the user input an character instead of a int. – Hero Man Aug 08 '18 at 09:50
@VeselinDavidov IT WORK THANK YOU VERY MUCH but can you explain it to me how the codes work? – Hero Man Aug 08 '18 at 09:52
added it in the answer – Veselin Davidov Aug 08 '18 at 10:30

INVALID INPUT AND LOOP

1 Answers1