Bash script: remove extension from file name [duplicate]

Question

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

The best solution I found so far is:

NAME=`echo "$FILE" | cut -d'.' -f1` EXTENSION=`echo "$FILE" | cut -d'.' -f2` 

This is wrong because it doesn't work if the file name contains multiple . characters. If, let's say, I have a.b.js, it will consider a and b.js, instead of a.b and js.

It can be easily done in Python with

file, ext = os.path.splitext(path) 

but I'd prefer not to fire up a Python interpreter just for this, if possible.

Any better ideas?

Answer 1

First, get file name without the path:

filename=$(basename -- "$fullfile") extension="${filename##*.}" filename="${filename%.*}" 

Alternatively, you can focus on the last '/' of the path instead of the '.' which should work even if you have unpredictable file extensions:

filename="${fullfile##*/}" 

You may want to check the documentation :

• On the web at section "3.5.3 Shell Parameter Expansion"
• In the bash manpage at section called "Parameter Expansion"

Answer 2

~% FILE="example.tar.gz"  ~% echo "${FILE%%.*}" example  ~% echo "${FILE%.*}" example.tar  ~% echo "${FILE#*.}" tar.gz  ~% echo "${FILE##*.}" gz 

For more details, see shell parameter expansion in the Bash manual.