Vtyjkyuk

I have a file that has a list of paths like so:

"1" "/user/bin/share"
"2" "/home/user/.local"
"3" "/root/"

Is there a way to extract just the paths? I dont want the numbers or quotation marks. How can I sed or grep the paths out of the file? What regex would be required for such a task?

If all the paths start with /, you could just match / followed by a sequence of non-" characters:

$ grep -o '/[^"]*' file
/user/bin/share
/home/user/.local
/root/

Alternatively, for a more structured approach, use awk to strip quotes from and print just the second field:

awk 'gsub(/"/,"",$2); print $2' file

Assuming that the paths do not contain newline characters,

$ sed 's/^.*[[:blank:]]"//; s/"$//' <file
/user/bin/share
/home/user/.local
/root/

The sed code first removes everything on each line up to and including the first " character preceded by a blank (space or tab). It then removes the " at the end.

This allows the paths to contain spaces and embedded " characters, but not the combination blank+".

Why wouldn't a simple

awk -F" 'print $4' file
/user/bin/share
/home/user/.local
/root/

work?

Simple approach. Pluck out field #4, with quotes as delimiters, using cut.

$ cut -f4 -d" file
/user/bin/share
/home/user/.local
/root/
$