Java While-Loop Program

Question

I am trying to get a valid input of "y", "Y", "n" , or "N".

If the input is not valid (for example any word that starts with a "y" or "n") I want it to re-prompt the user for input.

So far I have:

while  (again.charAt(0) != 'N' && again.charAt(0) != 'n' && again.charAt(0) !='Y' && again.charAt(0) != 'y' ) {
    System.out.println ("Invalid Inpur! Enter Y/N");
    again = numscan.next();
}

if (again.charAt(0)== 'N' || again.charAt(0) == 'n') {
    active = false;                   
} else {
    if (again.charAt(0)== 'Y' || again.charAt(0) == 'y'){
        active = true;
        random = (int) (Math.random () *(11));
    }
}

The problem I am having is if I enter any word that starts with the letter "y" or "n" it senses it as valid input (since it is the character at slot 0). I need help fixing this so I can re-prompt the user when they enter a word that starts with a "y" or "n".

Thanks!

Greg Hewgill · Accepted Answer · 2012-10-08 21:49:52Z

3

Assuming again is a string containing the complete user input, you could use:

while (!again.equals("N") && !again.equals("n") ...

The .equals() method will match only if the entire string matches.

answered Oct 8, 2012 at 21:49

Greg Hewgill

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1 Comment

arshajii Over a year ago

Remember that there is an equalsIgnoreCase method which could consolidate the condition of the while-loop.

arshajii · Accepted Answer · 2012-10-08 21:54:48Z

3

You could just test to make sure the length of the input is 1:

again.length() == 1

But a better approach might be:

while (! (again.equalsIgnoreCase("n") || again.equalsIgnoreCase("y"))) {
    ...
}

or even

while (! again.matches("[nyNY]")) {
    ...
}

edited Oct 8, 2012 at 21:54

answered Oct 8, 2012 at 21:48

arshajii

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Comments

kosa · Accepted Answer · 2012-10-08 21:48:47Z

1

One of the way would be:

First check again String length is only ONE character. If not, ask again.

if(again.length() ==1)
{
 while  (again.charAt(0) != 'N' && again.charAt(0) != 'n' && again.charAt(0) !='Y' && again.charAt(0) != 'y' ) {
               System.out.println ("Invalid Inpur! Enter Y/N");
                again = numscan.next();
            }
.....

}else
     {
System.out.println ("Invalid Inpur! Enter Y/N");
                    again = numscan.next();
    }

answered Oct 8, 2012 at 21:48

kosa

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Comments

CrazyCasta · Accepted Answer · 2012-10-08 21:49:47Z

1

It sounds like what you want is:

while  (!again.equals("N") && !again.equals("n") && !again.equals("Y") && !again.equals("y") ) {
    System.out.println ("Invalid Inpur! Enter Y/N");
    again = numscan.next();
}

This way you can also easily add Yes/No, etc later if you want.

answered Oct 8, 2012 at 21:49

CrazyCasta

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Comments

angel_007 · Accepted Answer · 2012-10-08 22:09:56Z

0

Regex could be an alternative to have strict input checks. Following piece of code validates y or n ignoring the case.

while (!again.matches("(?i)^[yn]$")){
    System.out.println ("Invalid Inpur! Enter Y/N");
    again = numscan.next();
}
active = (again.equalsIgnoreCase("Y"))? true : false;

answered Oct 8, 2012 at 22:09

angel_007

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