How do I output only a capture group with sed

Using sed:

sed -nE 's/^(pytest[^=]*=[^[:blank:]]*).*/\1/p' file

pytest==6.2.5
pytest-sugar==0.9.4
pytest-django==4.4.0

However a grep -o solution would be even simpler:

grep -o '^pytest[^=]*=[^[:blank:]]*' file

pytest==6.2.5
pytest-sugar==0.9.4
pytest-django==4.4.0

Explanation:

• ^pytest: Match pytest at the start
• [^=]*: Match 0 or more of any character except =
• =: Match a =
• [^[:blank:]]*: Match 0 or more of non-whitespace characters

You are missing the regex after #. This should solve it:

$ sed -nE "s/(^pytest.+)#.*/\1/p" ./requirements/local.txt

1st solution: With awk you could try following. Using match function of awk here, written and tested in GNU awk should work in any any. Simple explanation would be, using match function of awk to match regex ^pytest[^ ]* to match starting value of pytest till 1st occurrence of space and print the matched value by using substr function of awk.

awk 'match($0,/^pytest[^ ]*/){print substr($0,RSTART,RLENGTH)}' Input_file

2nd solution: Using GNU awk try following where making use of RS variable of it.

awk -v RS='(^|\n)pytest[^ ]*' 'RT{sub(/^\n*/,"",RT);print RT}' Input_file

As an alternative using awk, you might also set the field separator to # preceded by optional spaces, and print the first column if it starts with pytest

awk -F"[[:blank:]]*#" '/^pytest/ {print $1}' ./requirements/local.txt

Output

pytest==6.2.5
pytest-sugar==0.9.4
pytest-django==4.4.0

If the # is not always present, you could also make the match more specific to match the number, and then print the first field:

awk '/^pytest[^[:blank:]]*==[0-9]+(\.[0-9]+)*/ {print $1}' file