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

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]"
[download]

There are two sets of [/]-delimited strings in there. Without the "?" in the regular expression, perl would match from the first [ to the last ], putting the following into $1:

    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.