Regex matching/compiling question

[David Kaplowitz]

Hello,

Basically what I want to do is to take a string, and see if there's a single non-letter character return a boolean value.

Pattern p = Pattern.compile("[A-Za-z]");  // I'd really like to do ![A-Za-z]
Matcher m = p.matcher(name);
boolean matchFound = m.find();

But I'm only getting a false when all chars are non-letter. Should I be iterating the String in a loop and doing a comparison of the chars, or is there a way to do it with the regexp and I'm just not seeing how to do it?

Thanks for any hints,

Dave

[Mark Vedder]

To match a character that is not in a set of characters, you can use the ^ symbol (a circumflex accent) to mean 'not'. So [^A-Za-z] means find a character that is not an A-Z. You could also do [^A-Z] and set the case insensitive flag:

Pattern p = Pattern.compile("[^A-Z]", Pattern.CASE_INSENSITIVE);

That should work for what you want to do. Keep in mind this will only work on basic English letters. Letters from other languages, including accented letters will match the above pattern and be seen as a non-letter. Accented Characters are seen in English some times such as the word R�sum�.

p.s. www.regular-expressions.info has a great regex tutorial that might help you.
[ December 06, 2008: Message edited by: Mark Vedder ]

[David Kaplowitz]

Thanks a lot, Mark. That info helped get me on track with the pattern match. I'll check out that tutorial you linked to.

Best,

Dave

[Campbell Ritchie]

There is another good tutorial in the Java Tutorials.