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

Question

Hi all, I looked around for a while, but probably I can't "Google" with the proper keywords, so I'm asking here. I need to extract from a string a set of characters which are included between two delimiters, without returning the delimiters themselves. I'm trying to do it with C# RegEx object.

A simple example should be helpful:

Target: extract the substring between square brackets, without returning the brackets themselves.

Base string: This is a test string [more or less]

If I use the following reg. ex.

\[.*?\]

The match is "[more or less]". I need to get only "more or less", without the brackets. Is it possible to do it? Many thanks.

Answer 1

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.

Answer 2

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.

Answer 3

PHP:

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

Answer 4

I use the following site to test my Regular Expressions:

http://www.gskinner.com/RegExr/

Have you tried the following:

([A-Za-z])