Regex capture value within bracket, value contain brackets as well

Question

I want to match the following text:

(valuex) AnySingleWord (valuey)

I want to capture the text of valuex,valuey within the brackets. The tough thing is valuex,valuey both can contain brackets as well. For example

((value1) AnySingleWord (value2)) AnySingleWord ((value3) AnySingleWord (value4))

Then my valuex should be

(value1) AnySingleWord (value2)

and valuey should be

(value3) AnySingleWord (value4)

Can regex do something like counting the number of open bracket has passed, and match things until the right close bracket found? many thanks.

PS: I can limit number of open/close bracket pairs to 2-3, if this is small enough to be solvable by Regex

Related: [Can regular expressions be used to match nested patterns?](http://stackoverflow.com/q/133601/335858). — Sergey Kalinichenko, Jun 20 '13 at 18:16
Do you have a maximum number of nested brackets? If so, then you can use regex. However, it's probably not the easiest solution. — Jerry, Jun 20 '13 at 18:16
I can limit number of open/close bracket pairs to 2-3, if this is small enough to be solvable by Regex — Ben.Yan, Jun 20 '13 at 18:22
Headline change request. Shouldn't the word 'bracket' be used for '[' and not a '(' (parenthesis)? — crokusek, Mar 13 '14 at 18:13

score 2 · Accepted Answer · answered Jun 20 '13 at 18:30

2

Yes, you can use this pattern:

\(((?>\((?<DEPTH>)|\)(?<-DEPTH>)|[^()]+)*(?(DEPTH)(?!)))\)

more informations about this kind of pattern here:

http://blogs.msdn.com/b/bclteam/archive/2005/03/15/396452.aspx

answered Jun 20 '13 at 18:30

Casimir et Hippolyte

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This works, many thanks. Great as this is my individual work, otherwise teammate is gonna murder me, or I am gonna frustrating explain Regex balanced matching group. – Ben.Yan Jun 21 '13 at 20:56

score 1 · Answer 2 · answered Jun 20 '13 at 18:24

1

A pretty simple way would be as follows if I'm understanding you correctly:

1. iterate through characters
2. find first left brace
3. count braces until non-paired right brace appears and store index
4. take the substring from zero to that index and set as value x
5. delete value x substring
6. go to step 1 and store next part as value y

answered Jun 20 '13 at 18:24

eatonphil

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this would be intuitive a non-regex solution, if i can't manage to handle using regex, i would try this – Ben.Yan Jun 20 '13 at 19:04
Sure! It would also be a lot easier to read and debug. It's your choice, but this scenario certainly seems to lend itself to a non-regex solution. – eatonphil Jun 20 '13 at 19:09

score 1 · Answer 3 · answered Jun 20 '13 at 18:31

1

Your regex would be

(?<left>\(([^()]*(\([^()]*\))?[^()]*)+\))[^()]*(?<right>\([^()]*((\([^()]*\))?[^()]*)+\))

Your code would be

Match m=Regex.Match(inp,reg);
m.Groups["left"].Value;//left value
m.Groups["right"].Value;//rightvalue

answered Jun 20 '13 at 18:31

Anirudha

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Thanks, sadly Casimir's solution is shorter, thanks for the help again – Ben.Yan Jun 21 '13 at 20:58

Regex capture value within bracket, value contain brackets as well

3 Answers3