prevent sed from printing non-captured group

Question

I am using sed to capture and print only specific timestamps which are before 12pm (i.e. having 0 as first digit eg 09 as hours)

sed  -E 's/^.*\[(.* 0.*)\].*/\1/g'

However, it's also printing lines where no group is captured. How to avoid printing those?

Following is output of above command. Where regex matched, only captured group is printed and where it didn't match, original string is printed. I want to avoid printing string if regex is not matched.

2021/03/01 09:11:14
Accessed ([2021/03/01 12:32:36])
Accessed ([2021/03/01 16:48:29])
Accessed ([2021/03/01 19:40:03])
2021/03/02 08:53:27
Accessed ([2021/03/02 11:03:23])
Accessed ([2021/03/02 11:04:08])
Accessed ([2021/03/02 16:36:48])
Accessed ([2021/03/02 19:35:31])
2021/03/03 08:38:10

Original strings are there. Where regex matched, only capture group is printed and where it's not matched, full string is printed. Original string is Accessed ([2021/03/02 11:03:23]) — Shashwat Kumar, Mar 04 '21 at 07:31
It didn't print anything because given input has no match for space followed by `0` inside `[...]` — anubhava, Mar 04 '21 at 07:39
@anubhava If you could post your solution as an answer, then I can accept that. — Shashwat Kumar, Mar 04 '21 at 07:57
I beg to differ with the dupe post. There may be other more appropriate dupe that I could not find but this one doesn't look like a correct dupe. — anubhava, Mar 04 '21 at 14:36

anubhava · Accepted Answer · 2021-03-04T08:33:42.363

Converting my comment to answer so that solution is easy to find for future visitors.

You may use this sed to disable normal printing of unmatched lines:

sed -nE 's/^.*\[(.* 0.*)\].*/\1/p' file

Also please understand that .* is greedy in nature and due to lot of backtracking this pattern tends to get slower for large files.

I suggest using this regex with negated character class:

sed -nE 's/^[^[]*\[([^ ]* 0[^]]*)\].*/\1/p' file

prevent sed from printing non-captured group

1 Answers1