CODEWITHSUNDEEP
javaregex
artem
artem

Reputation: 16777

Java regexp - "()" brackets

Pattern pattern = Pattern.compile("<a>([a-zA-Z]+)</a>")
Matcher matcher = pattern.matcher("<a>Text</a>");
matcher.find()
String str = matcher.group();

I want to get "Text" to str, but I get "<a>Text</a>". Why and how should I do it properly?

Upvotes: 3

Views: 301

Answers (6)

collapsar
collapsar

Reputation: 17238

you have to call matcher.group with the number of your capture group - if you omit the argument the complete match will be returned.

best regards, carsten

ps: the best address to quickly solve these kinds of question is to look up the repective part of the java api docs.

Upvotes: 2

Kilian Foth
Kilian Foth

Reputation: 14366

group() returns the entire matched text. You want group(1), which returns the first paren-delimited group within the match. See the API docs.

Upvotes: 2

Dave Newton
Dave Newton

Reputation: 160261

You want group(1); the first group is the entire pattern.

See the group() and group(int) docs.

Upvotes: 2

Gareth Davis
Gareth Davis

Reputation: 28069

There is another overload of group() on matcher. Try:

matcher.group(1);

Upvotes: 2

Ry-
Ry-

Reputation: 225064

You need to specify the index of the group, 1 in this case:

Pattern pattern = Pattern.compile("<a>([a-zA-Z]+)</a>")
Matcher matcher = pattern.matcher("<a>Text</a>");
matcher.find()
String str = matcher.group(1);

Documentation for Matcher.group(int)

Upvotes: 4

ruakh
ruakh

Reputation: 183456

matcher.group(), with no arguments, returns the entire matched substring. Use matcher.group(1) to retrieve just the contents of the first parenthesized capture-group:

Pattern pattern = Pattern.compile("<a>([a-zA-Z]+)</a>")
Matcher matcher = pattern.matcher("<a>Text</a>");
matcher.find();
String str = matcher.group(1);

Upvotes: 5

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