Hide part of match from grep output

Instead of using extended-regex grep (-E), use perl-regex grep instead (-P), with a lookbehind and lookahead.

$ grep -oP "(?<=\[)[0-9].+(?=\])" logfile
113a6d9e-7b06-42c6-a52b-7a4e4d2e216c
113a6d9e-7b06-42c6-a52b-7a4e4d2e216c

Here, (?<=\[) indicates that there should be a preceding \[, and (?=\]) indicates that there should be a following \], but not to include them in the match output.

$ cat a.txt
test hello..[113a6d9e-7b06-42c6-a52b-7a4e4d2e216c]... this is
te [113a6d9e-7b06-42c6-a52b-7a4e4d2e216c].  this  is hello

$ grep -oP '(?<=\[)[^\]]*' a.txt
113a6d9e-7b06-42c6-a52b-7a4e4d2e216c
113a6d9e-7b06-42c6-a52b-7a4e4d2e216c

https://stackoverflow.com/a/19242713/6947646

sed is more applicable for the case than grep

sed '/\n/{P;D;};s/\[/\n/;s/\]/\n/;D' log