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.'-\\]*$");
Matcher m = pattern.matcher(testString);
I have to use the unicode value for some because I'm working with xhtml.
Any help would be great!
Assuming that you want to match \ and - and not ]:
Pattern pattern = Pattern.compile("^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'\\\\-]*$");
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.
It is hard to say what are you trying to achieve, but I can see few strange things in your regex:
\\] which makes ] normal character. 
] in your characters class then you need additional ] at the end, like "^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'-\\]]*$"
\ 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- 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 \\-
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