bash regex not working

Question

So I have this code

function test(){
  local output="ASD[test]"
  if [[ "$output" =~ ASD\[(.*?)\] ]]; then
     echo "success";
  else
     echo "fail"
  fi;
}

And as you can see it's supposed to echo success since the string matches that regular expression. However this ends up returning fail. What did I do wrong?

Yes it won't work because your regular expression is also not correct. What you actually want? — Manish, Apr 24 '16 at 03:53
@Manish regex is correct as well as code is running..don't understand OP's problem — rock321987, Apr 24 '16 at 03:54
it's seriously not working on my bash...I'm using version 4.1.17(0)-release — etherealeden, Apr 24 '16 at 03:57
@rock321987 i m not saying that regex is wrong but it won't match the value of $output. U got my point right. — Manish, Apr 24 '16 at 03:58
@Manish why isn't it matching....this guy says it matches: http://regexr.com/3d9i4 — etherealeden, Apr 24 '16 at 04:00
the terminal simply outputs "fail" ... nothing more nothing less, i've tried restarting bash, rerunning "source .bashrc" etc — etherealeden, Apr 24 '16 at 04:01
how its possible..its working everywhere..though the version of bash i am using is `4.3.11` — rock321987, Apr 24 '16 at 04:03
not sure if it matters but I'm also using the built in bash in this guy: http://mobaxterm.mobatek.net/ — etherealeden, Apr 24 '16 at 04:03
@rock321987 try using the mobaxterm terminal in that mobaxterm.mobatek.net link — etherealeden, Apr 24 '16 at 04:03
i think the problem can be with `\\`` because windows uses it in its path..try matching only `ASD` — rock321987, Apr 24 '16 at 04:04
is there any online version of that available or I need to install it? — rock321987, Apr 24 '16 at 04:07
You should rename your function `test()` not to confuse with the `test` shell builtin. — SLePort, Apr 24 '16 at 04:12
You need to escape `?` try `ASD\[.*\\?\]`..i am searching for its reason — rock321987, Apr 24 '16 at 04:27
I found that bash does not supports lazy match using `.*?`. So you have to use alternative approach suggested in other answer — rock321987, Apr 24 '16 at 04:53
Here, the solution is easy. If the delimiters were not single but multiple characters, I'd recommend [something like this](http://stackoverflow.com/questions/34311692/multiple-lazy-string-replacing-between-two-patterns-with-sed/34311727#34311727). — Wiktor Stribiżew, Apr 24 '16 at 07:13

Mark Reed · Accepted Answer · 2016-04-24T13:52:44.163

The ? in ASD\[(.*?)\] doesn't belong there. It looks like you're trying to apply a non-greedy modifier to the *, which is *? in Perl-compatible syntax, but Bash doesn't support that. (See the guide here.) In fact, if you examine $? after the test, you'll see that it's not 1 (the normal "string didn't match" result) but 2, which indicates a syntax error in the regular expression.

If you use the simpler pattern ASD\[(.*)\], then the match will succeed. However, if you use that regex on a string which might have later instances of brackets, too much will get captured by the parentheses. For example:

output=ASD[test1],ASD[test2]
[[ $output =~ ASD\[(.*)\] ]] && echo "first subscript is '${BASH_REMATCH[1]}'"
#=> first subscript is 'test1],ASD[test2'

In languages that support the *? syntax, it makes the matching "non-greedy" so that it will match the smallest string it can that makes the overall match succeed; without the ?, such expressions always match the longest possible instead. Since Bash doesn't have non-greediness, your best bet is probably to use a character class that matches everything except a close bracket, making it impossible for the match to move past the first one:

[[ $output =~ ASD\[([^]]*)\] ]] && echo "first subscript is '${BASH_REMATCH[1]}'"
#=> first subscript is 'test1'

Note that this breaks if there are any nested layers of bracket pairs within the subscript brackets - but then, so does the *? version.

bash regex not working

1 Answers1