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I am a relative beginner to Java and am having an issue with a user controlled do-while loop that accepts user input to repeat. Here is the code at the end of the loop. 

System.out.print("Do you wish to continue? (Y for yes " +
                            "/ N for no)");
        input = keyboard.nextLine();
        input.length();
        repeat = input.charAt(0);

        }while (repeat == 'Y' | repeat == 'y');

I know it's throwing the exception because of the value, but can't figure out what to do to fix it, as I'm sure it's relatively simple. Thanks for the help.

Edit 1: Stacktrace:

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: 
String index out of range: 0 
at java.lang.String.charAt(Unknown Source) 
at FractionTest.main(FractionTest.java:59)
Some guy
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Steve
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    Looks like you read empty line. – Pshemo Jul 22 '15 at 21:28
  • Sorry I left the stacktrace out. Here it is: Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 0 at java.lang.String.charAt(Unknown Source) at FractionTest.main(FractionTest.java:59) – Steve Jul 22 '15 at 21:30
  • possible duplicate of [IndexOutOfBoundsException when taking character input from the user](http://stackoverflow.com/questions/27619851/indexoutofboundsexception-when-taking-character-input-from-the-user) – naXa stands with Ukraine Jul 22 '15 at 21:35
  • Post stacktrace inside question. It is unreadable as comment. To do so use [edit] option. – Pshemo Jul 22 '15 at 21:43

2 Answers2

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It looks like you read empty line which returned you empty string "" so there is no characters there (not even 0th).

It usually happens when you are using nextLine right after other nextABC methods like nextInt since such methods doesn't consume line separators, and nextLine reads text until next line separator (or end of stream).

In that case you can add nextLine after that nextInt to consume line separator.

Anyway to avoid reading character from empty string and exception you can use something like 

} while (input.toLowerCase().startsWith("y"));

instead of 

    input.length();//this actually doesn't change anything, you can remove it
    repeat = input.charAt(0);

} while (repeat == 'Y' | repeat == 'y');
Pshemo
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  • I added the above if statement and it's still throwing the same exception. – Steve Jul 22 '15 at 21:39
  • @Steve I updated my answer to add alternative way of skipping this error, and probable cause of getting empty string. – Pshemo Jul 22 '15 at 21:46
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As @Phesmo mentioned you may be getting a 0-length line. This is an obscure problem. Have you called nextInt(), nextLong(), nextShort(), nextByte(), or nextBoolean() on the Scanner before calling nextLine()?

Try this

System.out.print("Do you wish to continue? (Y for yes / N for no)");

char ans;
do {
   ans = keyboard.next().charAt(0);
   // filter input
} while (ans == 'Y' || ans == 'y' || ans == 'N' || ans == 'n');

repeat = ans;

} while (repeat == 'Y' | repeat == 'y');

From documentation:

String Scanner#next()
Finds and returns the next complete token from this scanner. A complete token is preceded and followed by input that matches the delimiter pattern.

Or this

System.out.print("Do you wish to continue? (Y for yes / N for no)");    
repeat = keyboard.next("[YyNn]").charAt(0);

} while (repeat == 'Y' | repeat == 'y');

From documentation:

String Scanner#next(String pattern)
Returns the next token if it matches the pattern constructed from the specified string.

"[YyNn]" is a character class pattern, it means match any character from the set.

naXa stands with Ukraine
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