Regular Expression to find a string included between two characters while EXCLUDING the delimiters

Easy done:

(?<=\[)(.*?)(?=\])

Technically that's using lookaheads and lookbehinds. See Lookahead and Lookbehind Zero-Width Assertions. The pattern consists of:

is preceded by a [ that is not captured (lookbehind);

a non-greedy captured group. It's non-greedy to stop at the first ]; and

is followed by a ] that is not captured (lookahead).

Alternatively you can just capture what's between the square brackets:

\[(.*?)\]

and return the first captured group instead of the entire match.

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If you are using JavaScript, the solution provided by cletus, (?<=\[)(.*?)(?=\]) won't work because JavaScript doesn't support the lookbehind operator.

Edit: actually, now (ES2018) it's possible to use the lookbehind operator. Just add / to define the regex string, like this:

var regex = /(?<=\[)(.*?)(?=\])/;

Old answer:

Solution:

var regex = /\[(.*?)\]/;
var strToMatch = "This is a test string [more or less]";
var matched = regex.exec(strToMatch);

It will return:

["[more or less]", "more or less"]

So, what you need is the second value. Use:

var matched = regex.exec(strToMatch)[1];

To return:

"more or less"

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You just need to 'capture' the bit between the brackets.

\[(.*?)\]

To capture you put it inside parentheses. You do not say which language this is using. In Perl for example, you would access this using the $1 variable.

my $string ='This is the match [more or less]';
$string =~ /\[(.*?)\]/;
print "match:$1\n";

Other languages will have different mechanisms. C#, for example, uses the Match collection class, I believe.

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Here's a general example with obvious delimiters (X and Y):

(?<=X)(.*?)(?=Y)

Here it's used to find the string between X and Y. Rubular example here, or see image:

https://i.stack.imgur.com/jE62L.png

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[^\[] Match any character that is not [.

+ Match 1 or more of the anything that is not [. Creates groups of these matches.

(?=\]) Positive lookahead ]. Matches a group ending with ] without including it in the result.

Done.

[^\[]+(?=\])

Proof.

http://regexr.com/3gobr

Similar to the solution proposed by null. But the additional \] is not required. As an additional note, it appears \ is not required to escape the [ after the ^. For readability, I would leave it in.

Does not work in the situation in which the delimiters are identical. "more or less" for example.

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PHP:

$string ='This is the match [more or less]';
preg_match('#\[(.*)\]#', $string, $match);
var_dump($match[1]);

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Most updated solution

If you are using Javascript, the best solution that I came up with is using match instead of exec method. Then, iterate matches and remove the delimiters with the result of the first group using $1

const text = "This is a test string [more or less], [more] and [less]";
const regex = /\[(.*?)\]/gi;
const resultMatchGroup = text.match(regex); // [ '[more or less]', '[more]', '[less]' ]
const desiredRes = resultMatchGroup.map(match => match.replace(regex, "$1"))
console.log("desiredRes", desiredRes); // [ 'more or less', 'more', 'less' ]

As you can see, this is useful for multiple delimiters in the text as well

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This one specifically works for javascript's regular expression parser /[^[\]]+(?=])/g

just run this in the console

var regex = /[^[\]]+(?=])/g;
var str = "This is a test string [more or less]";
var match = regex.exec(str);
match;

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To remove also the [] use:

\[.+\]

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I had the same problem using regex with bash scripting. I used a 2-step solution using pipes with grep -o applying

 '\[(.*?)\]'  

first, then

'\b.*\b'

Obviously not as efficient at the other answers, but an alternative.

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I wanted to find a string between / and #, but # is sometimes optional. Here is the regex I use:

  (?<=\/)([^#]+)(?=#*)

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Here is how I got without '[' and ']' in C#:

var text = "This is a test string [more or less]";

// Getting only string between '[' and ']'
Regex regex = new Regex(@"\[(.+?)\]");
var matchGroups = regex.Matches(text);

for (int i = 0; i < matchGroups.Count; i++)
{
    Console.WriteLine(matchGroups[i].Groups[1]);
}

The output is:

more or less

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If you need extract the text without the brackets, you can use bash awk

echo " [hola mundo] " | awk -F'[][]' '{print $2}'

result:

hola mundo

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