Punctuation Regex in Java

Question

First, i\'m read the documentation as follow

http://download.oracle.com/javase/1.4.2/docs/api/java/util/regex/Pattern.html

And i want find any punctuation c

Answer 1

You may use character subtraction here:

String pat = "[\\p{Punct}&&[^@',&]]";

The whole pattern represents a character class, [...], that contains a \p{Punct} POSIX character class, the && intersection operator and [^...] negated character class.

A Unicode modifier might be necessary if you plan to also match all Unicode punctuation:

String pat = "(?U)[\\p{Punct}&&[^@',&]]";
              ^^^^

The pattern matches any punctuation (with \p{Punct}) except @, ', , and &.

If you need to exclude more characters, add them to the negated character class. Just remember to always escape -, \, ^, [ and ] inside a Java regex character class/set. E.g. adding a backslash and - might look like "[\\p{Punct}&&[^@',&\\\\-]]" or "[\\p{Punct}&&[^@',&\\-\\\\]]".

Java demo:

String value = "#`~!#$%^，";
String pattern = "(?U)[\\p{Punct}&&[^@',&]]";
Pattern r = Pattern.compile(pattern);    // Create a Pattern object
Matcher m = r.matcher(value);            // Now create matcher object.
while (m.find()) {
    System.out.println("Found value: " + m.group());
}

Output:

Found value: #
Found value: !
Found value: #
Found value: %
Found value: ，

Answer 2

You're matching two characters, not one. Using a (negative) lookahead should solve the task:

(?![@',&])\\p{Punct}