Question

I’m currently writing a bash script for renaming files.
Every file looks like this *_#.ext where * can be anything and # a 9 digits number. In this example, ill stick with 3 digits instead.

Here is a basic example of what it should do:

Input: example1_123.jpg
Output: example1.jpg

Simple enough? Right:

#!/bin/bash
filename="example1_123.jpg"
echo ${filename/_[0-9][0-9][0-9]/}

Output: example1.jpg

This works… as long as the input isn’t something like this filename:
example2_123.jpg_987.jpg

The file will be renamed to example2.jpg_987.jpg instead of example2_123.jpg.jpg

i tried using $ for end of line, but this breaks it, as $ is used for variables:

#!/bin/bash
filename="example2_123.jpg_987.jpg"
echo ${filename/_[0-9][0-9][0-9].jpg$/}.${filename/*./}

Output: example2_123.jpg_987.jpg.jpg
\$ also doesn’t work. I’m clueless…

Can anyone help me getting it to work the way I need it to?

P.S. [0-9]{3} instead of [0-9][0-9][0-9] also breaks it. If someone knows how to shorten it, please say so 🙂

Answer 1

Use:

target=$(echo $f | sed 's,_[0-9]\+\(\.[a-z]\+\)$,\1')

This will do what you want.

fg@erwin ~ $ f=example2_123.jpg_987.png
fg@erwin ~ $ target=$(echo $f | sed 's,_[0-9]\+\(\.[a-z]\+\)$,\1,')
fg@erwin ~ $ echo $target
example2_123.jpg.png
fg@erwin ~ $ f=example1_123.png
fg@erwin ~ $ target=$(echo $f | sed 's,_[0-9]\+\(\.[a-z]\+\)$,\1,')
fg@erwin ~ $ echo $target
example1.png

Surround $(...) with double quotes should you have space names in $f.

This is sed, therefore a classical regex dialect, therefore grouping, alternatives and quantifiers (well, except *) all need to be preceded with a backslash… The “canonical” regex really is:

_[0-9]+(\.[a-z]+)$