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I need a bash script to loop through a directory and do a filename change on each item: specifically on each file remove the last character prior to the extension. i.e. file1a.txt becomes file1.txt - The logic of accomplishing this escapes me. Or, is there a way to do this with mmv? The looping is no problem, it's zapping that last pre-extension character.

Thanks!

I can not help you with a shell script but I can help you do it in python script..

I will post this on my site as well.
You can check it out here http://linuxdynasty.org

#!/usr/bin/env python import os, re, sys os.chdir(sys.argv[1]) ls = os.listdir('./') for file_o in ls: file_n = re.sub("\w\.", ".", file_o) os.rename(file_o, file_n) print os.listdir("./")
[Modified by: linuxdynasty on April 28, 2008 11:26 PM]

[Modified by: linuxdynasty on April 28, 2008 11:43 PM]

[Modified by: linuxdynasty on April 28, 2008 11:44 PM]

[Modified by: linuxdynasty on April 29, 2008 12:58 PM]

As Larry Wall said, There's More Than One Way To Do It (you'll see it often as TMTOWTDI). So here's my version:

Some distributions have a "rename" utility. The problem is, it's not very consistent across distributions: some don't have it and some have a different syntax. Debian-based distributions (including Ubuntu) have the more powerful regular expression syntax, which you can use:

rename 's/(.*).\.(.+)/$1.$2/' *

Test with the "-n" option first to just see what would be done.

If you cannot have that, the simplest thing I could use is "sed". First verify if it works right, because the regular expression can get hairy:

ls | sed "s/\(.*\).\.\(.\+\)/mv \0 \1.\2/"

Then if you're sure of the result, execute it by piping to "sh":

ls | sed "s/\(.*\).\.\(.\+\)/mv \0 \1.\2/" | sh