Printing only text from group

Question

I have working example of substitution in online regex tester https://regex101.com/r/3FKdLL/1 and I want to use it as a substitution in sed editor.

echo "repo-2019-12-31-14-30-11.gz" | sed -r 's/^([\w-]+)-\d{4}-\d{2}-\d{2}-\d{2}-\d{2}-\d{2}.gz$.*/\1/p'

It always prints whole string: repo-2019-12-31-14-30-11.gz, but not matched group [\w-]+. I expect to get only text from group which is repo string in this example.

Answer 1

Try this:

echo "repo-2019-12-31-14-30-11.gz" |
sed -rn 's/^([A-Za-z]+)-[[:alnum:]]{4}-[[:digit:]]{2}-[[:digit:]]{2}-[[:digit:]]{2}-[[:digit:]]{2}-[[:digit:]]{2}.gz.*$/\1/p'

Explanations:

• \w will work (not [\w] wich matches either backslash or w), but you should use [[:alnum:]] which is POSIX
• For sed, \d isn't a regex class, but an escaped character representing a non-printable character
• Add -n to mute sed, with /p to explicitly print matched lines

---

Additionaly, you could refactor your regex by removing duplication:

echo "repo-2019-12-31-14-30-11.gz" |
sed -rn 's/^([[:alnum:]]+)-[[:digit:]]{4}(-[[:digit:]]{2}){5}.gz.*$/\1/p'

Answer 2

Looks like a job for GNU grep :

echo "repo-2019-12-31-14-30-11.gz" | grep -oP '^\K[[:alpha:]-]+'

Displays :

repo-

On this example :

echo "repo-repo-2019-12-31-14-30-11.gz" | grep -oP '^\K[[:alpha:]-]+'

Displays :

repo-repo-

Which I think is what you want because you tried with [\w-]+ on your regex.

If I'm wrong, just replace the grep command with : grep -oP '^\K\w+'