Regex: match words between certain characters, but not between other certain characters (in .NET)

Question

I want to match all the words in brackets, including brackets, which are not between simple quotation marks (in .NET):

"[field1] = 'id1'" -> match "[field1]"
"[field1] = '[field2]'" -> match "[field1]"
"[ field1 ] = '[ field2 ]'" -> match "[ field1 ]"
"[ field1 ] = '[ field2 ] field3'" -> match "[ field1 ]"
"[field1] = ' [field2] ' And [field3] = '[field4] '' [field5]'" -> match "[field1]" and "[field3]"

Any suggestion would be greatly appreciated!

Voting -1 should not be allowed without a commentting the reason. Anyway. — Szabolcs Antal, Mar 29 '17 at 11:57

score 0 · Answer 1 · edited May 23 '17 at 12:17

0

When you need to match things "which are not between" other things, you may think to negative lookaround:

(?<!' *)\[[^\]]*\](?! *')

The matched group is $0. If I am not wrong, .NET regexps allow you to use non-fixed width expressions inside lookaround (but be aware that it is very expensive), so the regex above should do the job.

In other languages this is not allowed, but there is a workaround called "possessive quantifier" (*+):

(?<!') *+(\[[^\]]*\]) *+(?!')

The matched group is $1. This works in all the languages in which negative lookaround and possessive quantifier are permitted (example). About the meaning of *+, I suggest you my answer on the topic.

But maybe the computationally simplest expression is:

[^'] *+(\[[^\]]*\]) *+[^']

The matched group is $1.

Update

This is a kind of heuristics:

(?<=^[^'\n\r]*(('[^'\n\r]*){2})*)\[[^\]]*\]

Or, maybe faster:

\[[^\]]*\](?=[^'\n\r]*(('[^'\n\r]*){2})*$)

The matched group is $0. Use it with multiline mode (m flag).

It will work if apexes inside the fields are not allowed.

edited May 23 '17 at 12:17

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answered Mar 29 '17 at 10:02

logi-kal

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Thanks a lot for your time, horcrux. I need something similar to the first regex, but it matches also the [field2] in: "[ field1 ] = '[field2] aaa'". Btw, is the "<" char needed at the beginning of the regex? I think it's not. – Szabolcs Antal Mar 29 '17 at 11:56
`(?<!A)` is a structure meaning "not preceded by `A`", so the character `<` is needed. – logi-kal Mar 29 '17 at 12:05
@ChocapicSz Anyway, what if you have `'[field1] [field2] '`? Should they be matched? – logi-kal Mar 29 '17 at 12:06
No, they shouldn't. – Szabolcs Antal Mar 29 '17 at 13:13
@ChocapicSz Ok. And do you think there will be some false positives or false negatives if we use the heuristic "The field is not preceded by an odd number of apexes"? – logi-kal Mar 29 '17 at 13:30
For an example of what I meant, see the answer update. – logi-kal Mar 29 '17 at 13:48
@ChocapicSz So what about the last regex I've appended? – logi-kal Mar 30 '17 at 06:45

Lőrinc Pap · Answer 2 · 2017-03-30T11:50:23.430

As @horcrux also pointed out, you can use negative lookbehind and lookahead. Since apparently C# supports infinite repetition, you could get away with:

"(?<!(')\s*)" + "\[[^\[\]]+\]" + "(?!\s*$1)

Separated the three parts, the negative lookbehind, the capturing match and the negative lookahead.

Note also that since the lookarounds are non-capturing, you don't even need parenthesis around the middle part.

In Java this would be something like:

"(?<!(')\s{0,1000})" + "\[[^\[\]]+\]" + "(?!\s{0,1000}$1)

but it's kinda' hacky, as it would only work for 1000 spaces between the two.

Regex: match words between certain characters, but not between other certain characters (in .NET)

2 Answers2