Extract filename and extension in Bash

Question

I want to get the filename (without extension) and the extension separately.

The best solution I found so far is:

NAME=`echo \"$FILE\" | cut -d\'.\'          


        

Answer 1

pax> echo a.b.js | sed 's/\.[^.]*$//'
a.b
pax> echo a.b.js | sed 's/^.*\.//'
js

works fine, so you can just use:

pax> FILE=a.b.js
pax> NAME=$(echo "$FILE" | sed 's/\.[^.]*$//')
pax> EXTENSION=$(echo "$FILE" | sed 's/^.*\.//')
pax> echo $NAME
a.b
pax> echo $EXTENSION
js

The commands, by the way, work as follows.

The command for NAME substitutes a "." character followed by any number of non-"." characters up to the end of the line, with nothing (i.e., it removes everything from the final "." to the end of the line, inclusive). This is basically a non-greedy substitution using regex trickery.

The command for EXTENSION substitutes a any number of characters followed by a "." character at the start of the line, with nothing (i.e., it removes everything from the start of the line to the final dot, inclusive). This is a greedy substitution which is the default action.