java regex pattern unclosed character class

Question

I need some help. Im getting:

Caused by: java.util.regex.PatternSyntaxException: Unclosed character class near index 24
^[a-zA-Z└- 0-9£µ /.'-\]*$
                        ^
        at java.util.regex.Pattern.error(Pattern.java:1713)
        at java.util.regex.Pattern.clazz(Pattern.java:2254)
        at java.util.regex.Pattern.sequence(Pattern.java:1818)
        at java.util.regex.Pattern.expr(Pattern.java:1752)
        at java.util.regex.Pattern.compile(Pattern.java:1460)
        at java.util.regex.Pattern.<init>(Pattern.java:1133)
        at java.util.regex.Pattern.compile(Pattern.java:823)

Here is my code:

String testString = value.toString();

Pattern pattern = Pattern.compile("^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'-\\]*$");
Matcher m = pattern.matcher(testString);

I have to use the unicode value for some because I'm working with xhtml.

Any help would be great!

Jeff · Accepted Answer · 2013-01-18 00:53:24Z

23

Assuming that you want to match \ and - and not ]:

Pattern pattern = Pattern.compile("^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'\\\\-]*$");

You need to double escape your backslashes, as \ is also an escape character in regex. Thus \\] escapes the backslash for java but not for regex. You need to add another java-escaped \ in order to regex-escape your second java-escaped \.

So \\\\ after java escaping becomes \\ which is then regex escaped to \.

Moving - to the end of the sequence means that it is used as a character, instead of a range operator as pointed out by Pshemo.

answered Jan 18, 2013 at 0:53

Jeff

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1 Comment

Joseph Vance Over a year ago

I think everyone hit on. I didn't realize I would need to quadruple '\' to achieve that backslash in my regex. I appreciate everyone's input.

Pshemo · Accepted Answer · 2013-01-18 00:53:53Z

2

It is hard to say what are you trying to achieve, but I can see few strange things in your regex:

you have opened class of characters but never closed it. Instead you used \\] which makes ] normal character.
- If you want to include ] in your characters class then you need additional ] at the end, like "^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'-\\]]*$"
- if you want to include \ in your characters class then you need to use \\\\ version, because you need to escape its special meaning two times, in regex engine, and in Javas String
you used - with ('-\\]) which in character class is used to specify range of characters like a-z or A-Z. To escape its special meaning you need to use \\-

edited Jan 18, 2013 at 0:53

answered Jan 18, 2013 at 0:46

Pshemo

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6 Comments

Jason Sperske Over a year ago

Could this work Pattern pattern = Pattern.compile("^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'\\-\\\\]*$"); ? Back slashes need to be quadruple escaped when written as Java Strings

Pshemo Over a year ago

@JasonSperske Good point with \\\\] in case OP wants to include \ in his characters class.

Jeff Over a year ago

@JasonSperske If you want to include - as a character, you can just put it at the end. If there's nothing after it in the character class it will be assumed not to be acting as an operator.

Pshemo Over a year ago

@Jeff from what I heard placing - at the end or start of character class works for Java but it might be not good practice since not all languages/regex engines accept this way of using it. So it is better to explicit escape its special meaning.

Joseph Vance Over a year ago

@Jeff correct. I wanted to include that so I kept it at the end.

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