my $string = "abe[123955785ada]sdjdjajd";
if ( $string =~ /\[(.*?)\]/ )
{
    my $inside = $1;
    print $inside, "\n";
}
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The string inside the brackets is in $inside.

The ( and ) around the .*? are "grouping parentheses"; they tell perl's regular expression engine to capture whatever it matches between them and store it in a special variable. The variable name will be a digit corresponding to the parenthesis group — $1 for the first group, $2 for the second group, etc. — counting '('s from the left.

So, in this case, perl captures what it finds between the [ and ] and stores that string in $1.

And what does it match? In this case, we've told it to match any character ("."), repeated 0 or more times ("*"), in a non-greedy manner ("?"). This last is best illustrated by example.

Say your original string was

    "aslkj[2099asgjskjw]asljgn[awoeiwj]"
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    2099asgjskjw]asljgn[awoeiwj
[download]

Most likely, this isn't what you want, so we put in the non-greedy modifier to make perl do what you want.

By the way, I said that "." matches any character; this isn't strictly true. It usually doesn't match carriage returns; if you want it to match carriage returns as well, add the /s modifier at the end of the regular expression.

For more information, see perlre. It explains all of this and more.

So, in this case, Perl captures what it finds between the [ and ] and stores that string in $1.

And what does it match? In this case, we've told it to match any character ("."), repeated 0 or more times ("*"), in a non-greedy manner ("?"). This last is best illustrated by example. Say your original string was

    aslkj[2099asgjskjw]asljgn[awoeiwj]
[download]

    2099asgjskjw]asljgn[awoeiwj
[download]

By the way, I said that "." matches any character; this isn't strictly true. It usually doesn't match carriage returns; if you want it to match carriage returns, add the "/s" modifier at the end of the regular expression.

For more information, take a look at perlre. It explains all of this and more.