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Renaming Groups of Files From the Command Line

Nov 18, 2009  By Mitch Frazier
 in
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Trying to figure how the command works

Anonymous's picture
Submitted by Anonymous (not verified) on Sat, 12/05/2009 - 11:44.

Enjoyed the video

Problem:
I can see the removal of the parts, but not the re-arrangement.

Background:
Ok, finally figured out the following:
1) The command uses "substring removal" / "parameter substitution". Took awhile to find this out - my first search term was " ${f# " and got stuff on F# prog language. Live and Learn.
2) The " f# " is for removing the shortest prefix (ie, from the left).
3) The " f$ " is for removing the shortest prefix (ie, from the right).

Problem Restated:
1) Where in the command does it say ----> move the " aa " from the the beginning to the end of the filename?

Correction to the video
1) " f# " is for removing the shortest _not_ the longest prefix

Link that was helpful: Linux Journal Article

I really enjoy learning about the CLI:) and look forward to these videos.

Zeek

Renaming Details

Mitch Frazier's picture
Submitted by Mitch Frazier on Sat, 12/05/2009 - 12:53.

First, if I said "longest" I mis-spoke: "#" and "%" remove the "shortest" match, "##" and "%%" remove the longest match.

The rename command is this: 

   mv -i $f 2009${f#*-2009}-${%-2009*}

There are 4 parts to the final file name: 

   2009                # Literal
   ${f#*-2009}         # Based on original name
   -                   # Literal
   ${f%-2009*}         # Based on original name

Given the name "aa-2009-01-01". The first substitution removes the shortest prefix that matches "*-2009", in this case that prefix is "aa-2009", leaving "-01-01". The second substitution removes the shortest suffix that matches "-2009*", in this case that suffix is "-2009-01-01", leaving "aa". So the final name is formed from: 

   2009                ==> 2009
   ${f#*-2009}         ==> -01-01
   -                   ==> -
   ${f%-2009*}         ==> aa

Resulting in the file name "2009-01-01-aa", the desired result.

Mitch Frazier is an Associate Editor for Linux Journal.

Thanks

Anonymous's picture
Submitted by Anonymous (not verified) on Sat, 12/05/2009 - 14:58.

I got it!
Very Clever
That was a surprisingly quick reply - Thanks

Zeek
Cheers

Ways

hemanth.hm's picture
Submitted by hemanth.hm (not verified) on Sat, 11/21/2009 - 01:04.

rename 's/foo/bar/g' *

rename .oldextension .newextension *.oldextension

mmv 'banana_*_*.asc' 'banana_#2_#1.asc'

Bulk renaming : http://is.gd/50bGB

You can also use "rename" command

apexwm's picture
Submitted by apexwm (not verified) on Thu, 11/19/2009 - 07:37.

The "rename" command offers as way to directly do essentially the same thing. It does a find/replace of the groups of files you specify.

Only in a few cases

Mitch Frazier's picture
Submitted by Mitch Frazier on Thu, 11/19/2009 - 09:20.

Rename will work for you in some cases but not all, specifically it wouldn't work for the example in the video, unless I'm missing something...

Mitch Frazier is an Associate Editor for Linux Journal.

rename will work with

jimi's picture
Submitted by jimi (not verified) on Mon, 01/11/2010 - 17:18.

rename will work with regex...

    rename s/\(..\)-\(.*\)/\$2-\$1/ *2009*

... but I never considered this usage of parameter substitution. Thanks!