Issue
I saw a bash command sed 's%^.*/%%'
Usually the common syntax for sed is sed 's/pattern/str/g', but in this one it used s%^.* for the s part in the 's/pattern/str/g'.
My questions:
What does s%^.* mean?
What's meaning of %% in the second part of sed 's%^.*/%%'?
Solution
The % is an alternative delimiter so that you don't need to escape the forward slash contained in the matching portion.
So if you were to write the same expression with / as a delimiter, it would look like:
sed 's/^.*\///'
which is also kind of difficult to read.
Either way, the expression will look for a forward slash in a line; if there is a forward slash, remove everything up to and including that slash.
Answered By - Mark Rushakoff Answer Checked By - Cary Denson (WPSolving Admin)