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

1
1906
7/10/2018 3:04:27 PM

Accepted Answer

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 :

3224
11/11/2018 5:41:47 AM

~% 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.


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