I would like to extract css comments from style.css.
Comments are in the beginning of the css files and they are formatted like this
/*
Author:name
URI:link
etc
*/
After some searches I found :
sed -n '/^\/\*\$,/\*\/$/p' style.css
It didn't work, I had this error :
sed: -e expression #1, char 11: unknown command: `\'
Do you suggest a good solution.
If possible with a good link for a tutorial for basic sed uses for beginners.
This sed should work:
sed -n '/\/\*/,/\*\//{P;/\*\//q;D;}' style.css
/* and */, your suggestion will show all the blocs. What should I add then ?/* and */ from the result, I was thinking that -n is enough but no, the patterns shows in the final result. I think I should add another sed to substitute them with blank.| sed '1d;$d'
sed -n '/\/\*/,/\*\//{/\*\//q;/\/\*/!P;D;}' style.css would also work same way.I think you mean:
sed -n '/^\/\*$/,/\*\/$/p' style.css
You had $ and / in the first regep backwards.
/ , but executing this will show nothing.sed -n '/^\/\*\$/,/\*\/$/p' style.css will not show anything
$, which makes it a literal instead of matching the end of line./* or */.If they are always in a colon delimited format like that, you can use:-
sed -n '/Author/p' | cut -d':' -f2
That will get you the author name. I'd definitely recommend learning about cut for colon delimited files.
