Regular Expressions on Punctuation

user_1163842 0
Viewed 121805

So I'm completely new to regular expressions, and I'm trying to use Java's java.util.regex to find punctuation in input strings. I won't know what kind of punctuation I might get ahead of time, except that (1) !, ?, ., ... are all valid puncutation, and (2) "<" and ">" mean something special, and don't count as punctuation. The program itself builds phrases pseudo-randomly, and I want to strip off the punctuation at the end of a sentence before it goes through the random process.

I can match entire words with any punctuation, but the matcher just gives me indexes for that word. In other words:

Pattern p = Pattern.compile("(.*\\!)*?");
Matcher m = p.matcher([some input string]);

will grab any words with a "!" on the end. For example:

String inputString = "It is a warm Summer day!";
Pattern p = Pattern.compile("(.*\\!)*?");
Matcher m = p.matcher(inputString);
String match = inputString.substring(m.start(), m.end());

results in --> String match ~ "day!"

But I want to have Matcher index just the "!", so I can just split it off.

I could probably make cases, and use String.substring(...) for each kind of punctuation I might get, but I'm hoping there's some mistake in my use of regular expressions to do this.

regex java string
4 Answers

I would try

\W

it matches any non-word character. This includes spaces and punctuation, but not underscores. It’s equivalent to [^A-Za-z0-9_]

user_87477660

I was tring to find how to replace a regex, with keeping other regex part. Example: Hi , how are you ? -> Hi, how are you?. After studying a little i found that i could create groups, using "()", so just replaced the goup one, that was "(\s)".

        String a = "Hi , how are you ?";
        String p = "(\s)([,.!?\\-])";
        System.out.println(a.replaceAll(p,"$2"));
        //output: Hi, how are you?
user_126472390
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