Regular Expression Java (punctuations)

Question

I would like to identify 2 types of patterns in a string. They are:

1) xxx:xxxxxxx:xxx.xx

2) xxx.xxx.xxx.xx:xxxxxxxxxx

Basically I want to know how to identify a literal "." in a string.

Since . means any character, what should I type when I am looking for a literal "."?

use the java Pattern and Matcher with \\. [source][1] [1]: http://stackoverflow.com/questions/3674930/java-regex-meta-character-and-ordinary-dot — grepit, May 17 '13 at 15:53

score 10 · Answer 1 · answered May 17 '13 at 15:51

10

You can either escape the . like this \\.

or

use it within character class like this [.]

answered May 17 '13 at 15:51

Anirudha

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score 0 · Answer 2 · answered May 17 '13 at 15:56

Try using, String [] stringArray = string.split("\\."); escaping the "."

And then int periods = stringArray.length;

Telling you how many periods there are in your "String"

Just a start on escaping characters. I am unsure what your question is asking so I just gave an intro to escaping.

score 0 · Answer 3 · answered May 17 '13 at 16:05

good luck

    String text    ="xxx.xxx.xxx.xx:xxxxxxxxxx";

    String patternString = "(.*)(\\.)(.*)";

    Pattern pattern = Pattern.compile(patternString);

    Matcher matcher = pattern.matcher(text);
    boolean matches = matcher.matches();
    System.out.println(matches);
}

Regular Expression Java (punctuations)

3 Answers3