I need to retrieve the filename for a bash script. I thought mc ls could do all that ls can do, but I seem to be mistaken. So now I'm struggling with regex.
When I do mc ls minio/bucket1/, I'll get:
[2021-05-14 11:15:18 CEST] 0B files1/
[2021-05-14 11:15:18 CEST] 0B files2/
[2021-05-14 11:15:19 CEST] 0B file1.ext
[2021-05-14 11:15:18 CEST] 0B file2.ext
How can I extract just the filename in bash?
You can pipe the following sed command after your mc ls command:
sed -n 's/^\[[^][]*][[:blank:]]*[^[:blank:]]*[[:blank:]]\(.*\.gpkg\)$/\1/p'
See the online demo.
Details:
-n - suppresses the default line outputs - substitution command^\[[^][]*][[:blank:]]*[^[:blank:]]*[[:blank:]]*\(.*\.gpkg\)$ - a regex that matches
^ - start of string\[[^][]*] - a substring between [ and ] with no square brackets inside[[:blank:]]* - zero or more horizontal whitespace[^[:blank:]]* - zero or more non-horizontal whitespace chars[[:blank:]]* - zero or more horizontal whitespace\(.*\.gpkg\) - Group 1: any text and then .gpkg$ - end of string\1 - replaces the match with Group 1 valuep - prints the result of the substitution.