Regular expression to match brackets

Question

For a templating engine, I am using regular expressions to identify content under brackets in a string. For example the regex needs to match {key} or <tag> or [element].

Currently my regular expression looks like this:

var rx=/([\[\{<])([\s\S]+?)([\]\}>])]/;

The issue is that such a regular expression doesn't force brackets to match. For example in the following string:

[{lastName},{firstName}]

the regular expression will match [{lastName}

Is there a way to define matching brackets? Saying for example that if the opening bracket is a [ then the closing bracket must be a ], not a } or a >

Remember that characters inside square bracket expressions often lose their specialness. Thus `[[]` matches a left-square bracket, `[{]` matches a left curly bracket, etc. — ghoti, Aug 10 '12 at 18:33
It sounds like what you need is a parser, not a regular expression. — gcochard, Aug 10 '12 at 18:33
do all of those brackets mean the same thing? If not, you just want different regexes — murgatroid99, Aug 10 '12 at 18:34
@ghoti correct, I just thought it was better for readability. — Christophe, Aug 10 '12 at 18:38
Um, no, actually, the mess of extra unnecessary characters makes the regex *less* readable. :) — ghoti, Aug 10 '12 at 18:40
@Greg right, but parsers are heavy and not adapted to my custom case. In the worst case... — Christophe, Aug 10 '12 at 18:41
...I'll use 3 different regexes, as suggested by @murgatroid99 — Christophe, Aug 10 '12 at 18:42
This is a [duplicate](http://stackoverflow.com/questions/546433/regular-expression-to-match-outer-brackets) but suffice it to say, no, you cannot do this. — Malvolio, Aug 10 '12 at 19:40

score 26 · Accepted Answer · answered Aug 10 '12 at 18:53

26

The best way to do this, especially if different brackets can have different meanings, is to split into 3 regular expressions:

var rx1 = /\[([^\]]+)]/;
var rx2 = /\(([^)]+)\)/;
var rx3 = /{([^}]+)}/;

These will match any text surrounded by [], (), and {} respectively, with the text inside in the first matched group.

answered Aug 10 '12 at 18:53

murgatroid99

15,284
8
52
88

(marking this one as answer as it was the first - in the comments) – Christophe Aug 11 '12 at 00:06

StrubT · Answer 2 · 2012-08-10T18:48:22.557

4

you could use alternatives using pipe character (|) like this one /\[([\s\S]+?)\]|\{([\s\S]+?)\}|<([\s\S]+?)>/, although it gets pretty long.

EDIT: shortend the regex, is not that long any more...

edited Aug 10 '12 at 18:48

answered Aug 10 '12 at 18:42

StrubT

998
7
18

3

Instead of `([\s\S]+?)` you could use something like `([^\]]+)`, which would make it clearer that you are matching the longest sequence of characters that does not contain the ending bracket (and shorter) – murgatroid99 Aug 10 '12 at 18:48

score 4 · Answer 3 · answered Jan 17 '13 at 00:15

4

var rx = /\[[^\]]+\]|\{[^}]+\}|<[^>]+>/;

answered Jan 17 '13 at 00:15

MikeM

9,855
2
27
42

right, this is similar to the first answer. I was actually interested in the use of an object you showed in another post: {"[":"]","",...}. The object used in a for...in loop might help make the pattern more generic. – Christophe Jan 17 '13 at 00:53
1

@Christophe. The answer you have accepted is of poor quality. You would have to go through the matches of the three regexs and compare them: e.g. with "{[abc]}" the first regex will match [abc] which is not what you want etc. Further, the answer doesn't handle `<>` at all and it handles `()` unnecessarily. You are not obliged to accept the first answer just because it is half-way correct - and the one you have accepted isn't even that! – MikeM Jan 17 '13 at 14:50
Well, it sounded like a good idea at the time... I am actually thinking that the solution I posted in http://stackoverflow.com/questions/14334740/missing-parentheses-with-regex/14369329#14369329 might be a good fit here too. – Christophe Jan 17 '13 at 19:08

ghoti · Answer 4 · 2012-08-10T19:03:42.010

Is there a way to define matching brackets? Saying for example that if the opening bracket is a [ then the closing bracket must be a ], not a } or a >

Sort-of.

ERE does not provide a way for you to match a closing bracket to an opening bracket the way you describe. (It may be possible using PREG magic, but I'll have to leave that for someone else.) You'll need either to have multiple regular expressions, or multiple atoms within a single regular expression.

If you use a single regex, I gather you'll need to determine the type of bracketed string you're detecting, as well as the content of that string. As was mentioned in comments, you'll need to do this in your programming language, but you can at least get what you need out of the regex.

In the regex below, each style of string is represented as a "branch" in the RE. Branches are separated by or-bars (|). For clarity, I'm assuming all strings are [:alnum:]. You haven't specified content, so you'll need to adjust for your particular requirements.

/(\[)([[:alnum:]]+)\]|(\()([[:alnum:]]+)\)|(\{)([[:alnum:]]+)\}/
 ↑   ↑               ↑                    ↑
 $1  $2           divider              divider

Note that in each branch, the first character is enclosed by round brackets, making it an "atom". You need your code to refer to this atom like a backreference. The second atom is the inner string. Now ... my JavaScript isn't as strong as my, say, baking skill, but this might be a start:

String.prototype.bracketstyle = function() {
  var re = /(\[)([:alnum:]+)\]|(\()([:alnum:]+)\)|(\{)([:alnum:]+)\}/;
  return this.replace(re,"$1");
}

String.prototype.innerstring = function() {
  var re = /(\[)([:alnum:]+)\]|(\()([:alnum:]+)\)|(\{)([:alnum:]+)\}/;
  return this.replace(re,"$2");
}

I suspect you could combine these into a single function, or use them differently without making them a function, but you get the idea.

This looks similar to the first answer. But I am afraid JavaScript doesn't understand :alnum: — Christophe, Aug 11 '12 at 00:01
Not much of a shortcut ;-) and right, it works in several languages, but not in JavaScript. — Christophe, Aug 13 '12 at 16:45
What I'm saying is that if `[[:alnum:]]+` doesn't work for you, simply replace it with `[A-Za-z0-9]+`. — ghoti, Aug 13 '12 at 20:50

Regular expression to match brackets

4 Answers4

Linked