Java - regex - group separation

Question

I need to match entries like {@x anything} in strings like a b c 1 225 {@x anything here1} test test {@x blabla} xyz test {@x any characters here}. I tried \{(@x ([^\}].*\w+(\.*)\s*)*)\} so far, but this is not really what I want and I am kinda stuck :(

So that should get:

anything here1

blabla

any characters here

So, what's the problem? What have you tried so far?

AlexR
– AlexR

2012-03-19 15:29:10 +00:00
Commented Mar 19, 2012 at 15:29 — AlexR
– AlexR, Commented Mar 19, 2012 at 15:29
I mentioned I tried .... so far

Display Name
– Display Name

2012-03-19 15:30:27 +00:00
Commented Mar 19, 2012 at 15:30 — Display Name
– Display Name, Commented Mar 19, 2012 at 15:30

Bogdan Emil Mariesan · Accepted Answer · 2012-03-19 15:41:27Z

2

Well to extract all groups that come with that structure you can start with:

{@x [a-zA-Z0-9 ]+}

From this point just remove the header and end of the requested strings and you should have the required output.

EDIT:

I've updated the regex a bit:

{@x [\w= ]+}

edited Mar 19, 2012 at 15:41

answered Mar 19, 2012 at 15:30

Bogdan Emil Mariesan

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2 Comments

Display Name Over a year ago

I might need to have special characters there, such as the equal (=) sign.

Sergiu Over a year ago

I think he says that he can have multiple special characters(not only the equal sign).

yggdraa · Accepted Answer · 2012-03-19 15:48:38Z

1

Try this one: "({@x ([^{]*)})"

   String string = "a b c 1 225 {@x anything = here1} test test {@x bl** #abla} xyz test {@x any characters here}";        
    String regexp = "(\\{\\@x ([^\\{]*)\\})";
    Pattern pattern = Pattern.compile(regexp);
    Matcher matcher = pattern.matcher(string);
    while (matcher.find()){
        System.out.println(matcher.group(2));
    }

edited Mar 19, 2012 at 15:48

answered Mar 19, 2012 at 15:42

yggdraa

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Comments

barsju · Accepted Answer · 2012-03-19 15:52:56Z

1

This should do it:

String string = "a b c 1 225 {@x anything here1} test test {@x blabla} xyz test {@x any characters here}";
String regexp = "\\{\\@x ([^\\}]*)\\}";
Pattern pattern = Pattern.compile(regexp);
Matcher matcher = pattern.matcher(string);
while (matcher.find()){
    System.out.println(matcher.group(1));
}

\{\@x - Matchers you start.

([^\}]*) - matches anyting except the end-curly (}) and puts that in a group (1)

\} - matches the end curly

Then you search and subtract your group.

answered Mar 19, 2012 at 15:52

barsju

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Comments

ring bearer · Accepted Answer · 2012-03-19 16:22:23Z

1

Another try:

  String input="a b c 1 225 {@x anything here1} test test {@x blabla} xyz test {@x any characters here}";
  String pattern = "\\{@x [(\\w*)(\\s*)]*\\}";
  for( String s: input.split(pattern)){
      System.out.println(s);
  }

\w* = any word ( a-z,A-Z,0-9) ; *= 0 or more

\s* = white space ; *= 0 or more

[]* - repeating group.

edited Mar 19, 2012 at 16:22

answered Mar 19, 2012 at 15:49

ring bearer

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Comments

Betlista · Accepted Answer · 2012-03-22 08:38:37Z

1

It seems to me that you want to tell in pattern "match the minimum between characters '{' and '}'" so the pattern you can use is:

    final String string = "a b c 1 225 {@x anything = here1} test test {@x bl** #abla} xyz test {@x any characters here}";
    final Pattern pattern = Pattern.compile( "\\{@x (.*?)\\}" ); // <-- pattern
    final Matcher matcher = pattern.matcher( string );
    while ( matcher.find() )
        System.out.println( matcher.group( 1 ) );

? in .*? is doing exactly that.

answered Mar 22, 2012 at 8:38

Betlista

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