code statement for validating an exit?

Question

am having trouble trying to validate a user response to exit my app or try again(its a simple game)? when the game finishes I ask the user if they want to continue type "y" for yes or "n" for no to exit. How can I validate their response so if its neither y or n I show an error message and ask them to type it again???

This doesn't seem to worK for me???

if (!choice.equalsIgnoreCase("n") && !choice.equalsIgnoreCase("y"))

           System.out.println("Error enter y for yes and n for no");
            System.out.println("Would you like to play again (y/n):");
           choice = sc.next();

You need to post your full code, so we can help you. This snippet doesn't explains what or where is the error. — João Vitor Verona Biazibetti
– João Vitor Verona Biazibetti, Commented Feb 24, 2013 at 3:21
no it was the logic, using an "if" statement doesn't end the program regardless of the code within, it needs to be in a "while" loop context for the code to work. — babaysteps
– babaysteps, Commented Feb 24, 2013 at 3:45
it was mearly a logic issues not a code issue that way i posted a snippet instead of my whole code. — babaysteps
– babaysteps, Commented Feb 24, 2013 at 3:48

LNO · Accepted Answer · 2013-02-24 03:28:19Z

2

Just change the

if (!choice.equalsIgnoreCase("n") && !choice.equalsIgnoreCase("y"))

to

while (!choice.equalsIgnoreCase("n") && !choice.equalsIgnoreCase("y"))

This will continue to ask them for another letter each time they don't choose the letter 'n' or 'y'.

EDIT: Also, you might want brackets around the code within the if (soon to be while) statement.

answered Feb 24, 2013 at 3:28

LNO

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Mik378 · Accepted Answer · 2013-02-24 03:24:55Z

0

Without any full context, difficult to give a relevant answer.

But you can already test your code with curly braces around the if statement:

if (!choice.equalsIgnoreCase("n") && !choice.equalsIgnoreCase("y")){
  System.out.println("Error enter y for yes and n for no");
  System.out.println("Would you like to play again (y/n):");
  choice = sc.next();
}

Indeed, with your current code, only the first line:

System.out.println("Error enter y for yes and n for no");

is taken in account by your condition.

answered Feb 24, 2013 at 3:24

Mik378

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