Hello,

I'm good with Unix, but I haven't learned Perl yet.... I know Perl can
do this, but since I don't know Perl yet, I'm a little hung up.

Problem:

I need to rename a batch of files:

foo_001.jpg
foo_002.jpg
foo_003.jpg
bar_001.jpg
bar_002.jpg
bar_003.jpg
.....and so on.

I need to take the number on these files (notice how the number is
duplicated on foo and bar) and make the entire batch consecutive. In
other words, I want the file list to look like this:

foo_001.jpg
foo_002.jpg
foo_003.jpg
bar_004.jpg
bar_005.jpg
bar_006.jpg
.....and so on.

I'd appreciate any ideas! I just feel like this is so simple and I'm
just not seeing the answer!

Thanks -

Peter

In article <1188589094.698861.37200@i38g2000prf.googlegroups. com>,
gibbonsp <gibbonsp@gmail.com> wrote:
(From)
>foo_001.jpg
>foo_002.jpg
>foo_003.jpg
>bar_001.jpg
>bar_002.jpg
>bar_003.jpg
>....and so on.

(To)

>foo_001.jpg
>foo_002.jpg
>foo_003.jpg
>bar_004.jpg
>bar_005.jpg
>bar_006.jpg
>....and so on.

gawk -F_ '{ print "mv -iv",$0,$1"_"sprintf("%03d.jpg",++n) }' file1 > /tmp/dofile

Then bring up /tmp/dofile in your editor, check it over, and when you
are happy with it (and you've confirmed that "mv -iv" behaves the same
way on your system as it does on mine), source the file from your shell.

2007-08-31, 19:38(-00), gibbonsp:
> Hello,
>
> I'm good with Unix, but I haven't learned Perl yet.... I know Perl can
> do this, but since I don't know Perl yet, I'm a little hung up.
>
> Problem:
>
> I need to rename a batch of files:
>
> foo_001.jpg
> foo_002.jpg
> foo_003.jpg
> bar_001.jpg
> bar_002.jpg
> bar_003.jpg
> ....and so on.
>
> I need to take the number on these files (notice how the number is
> duplicated on foo and bar) and make the entire batch consecutive. In
> other words, I want the file list to look like this:
>
> foo_001.jpg
> foo_002.jpg
> foo_003.jpg
> bar_004.jpg
> bar_005.jpg
> bar_006.jpg
> ....and so on.
[...]

zsh (assuming autoload -U zmv in ~/.zshrc):

n=0; zmv -n '(*_)<->(*)' '$1${(l:3::0$((++n))}$2'

recursively:

n=0; zmv -n '(**/)(*_)<->(*)' '$1$2${(l:3::0$((++n))}$3'

Remove "-n" when happy.

--
Stéphane

In article <1188589094.698861.37200@i38g2000prf.googlegroups. com>,
gibbonsp <gibbonsp@gmail.com> wrote:
>I need to rename a batch of files:
>
>foo_001.jpg
>foo_002.jpg
>foo_003.jpg
>bar_001.jpg
>bar_002.jpg
>bar_003.jpg
>....and so on.
>
>I need to take the number on these files (notice how the number is
>duplicated on foo and bar) and make the entire batch consecutive. In
>other words, I want the file list to look like this:
>
>foo_001.jpg
>foo_002.jpg
>foo_003.jpg
>bar_004.jpg
>bar_005.jpg
>bar_006.jpg
>....and so on.
>
>I'd appreciate any ideas! I just feel like this is so simple and I'm
>just not seeing the answer!

A ksh93 utility that has various options for just this sort of thing:

ftp://ftp.armory.com/pub/scripts/ren

In this case, it would be:

ren -m2=s -z3 '*_*.jpg'

("replace the second globbed segment with sequential integers that
contain 3 digits")

John
--
John DuBois spcecdt@armory.com KC6QKZ/AE http://www.armory.com/~spcecdt/

In article <1188589094.698861.37200@i38g2000prf.googlegroups. com>,
gibbonsp <gibbonsp@gmail.com> wrote:
>I need to rename a batch of files:
>
>foo_001.jpg
>foo_002.jpg
>foo_003.jpg
>bar_001.jpg
>bar_002.jpg
>bar_003.jpg
>....and so on.
>
>I need to take the number on these files (notice how the number is
>duplicated on foo and bar) and make the entire batch consecutive. In
>other words, I want the file list to look like this:
>
>foo_001.jpg
>foo_002.jpg
>foo_003.jpg
>bar_004.jpg
>bar_005.jpg
>bar_006.jpg
>....and so on.
>
>I'd appreciate any ideas! I just feel like this is so simple and I'm
>just not seeing the answer!

A ksh93 utility that has various options for just this sort of thing:

ftp://ftp.armory.com/pub/scripts/ren

In this case, it would be:

ren -m2=s -z3 '*_*.jpg'

("replace the second globbed segment with sequential integers that
contain 3 digits")

John
--
John DuBois spcecdt@armory.com KC6QKZ/AE http://www.armory.com/~spcecdt/

On Sep 2, 1:33 am, spce...@armory.com (John DuBois) wrote:
> In article <1188589094.698861.37...@i38g2000prf.googlegroups. com>,
>
>
>
> gibbonsp <gibbo...@gmail.com> wrote:
> >I need to rename a batch of files:
>
> >foo_001.jpg
> >foo_002.jpg
> >foo_003.jpg
> >bar_001.jpg
> >bar_002.jpg
> >bar_003.jpg
> >....and so on.
>
> >I need to take the number on these files (notice how the number is
> >duplicated on foo and bar) and make the entire batch consecutive. In
> >other words, I want the file list to look like this:
>
> >foo_001.jpg
> >foo_002.jpg
> >foo_003.jpg
> >bar_004.jpg
> >bar_005.jpg
> >bar_006.jpg
> >....and so on.
>
> >I'd appreciate any ideas! I just feel like this is so simple and I'm
> >just not seeing the answer!


Assuming that your filenames follow the pattern .*_¥d{3}.jpg


#!/bin/bash

#dir files contains all the files
files=`ls files`
i=1
padding="000"
for file in $files
do
((len=3-${#i}))
j="${padding:0:${len}}${i}"
newName=${file/_???./_${j}.}
mv files/${file} files/${newName}
((i=i+1))
done

arch119 wrote:
>
> Assuming that your filenames follow the pattern .*_¥d{3}.jpg
>
>
> #!/bin/bash
>
> [...]
> i=1
> padding="000"
> [...]
> ((len=3-${#i}))
> j="${padding:0:${len}}${i}"
> [...]

If you want leading zeros:

$ i=7
$ printf -v j "%03d" $i
$ echo $j
007

--
Best regards | "The only way to really learn scripting is to write
Cyrus | scripts." -- Advanced Bash-Scripting Guide

Hello,

arch119 wrote:
>
> #!/bin/bash
>
> #dir files contains all the files
> files=`ls files`

With output of "ls" you get problems with spaces and line breaks in filenames.

> i=1
> padding="000"
> for file in $files

better:
for file in *

> do
> ((len=3-${#i}))
> j="${padding:0:${len}}${i}"
> newName=${file/_???./_${j}.}
> mv files/${file} files/${newName}
> ((i=i+1))
> done



If you want leading zeros:

$ i=7
$ printf -v j "%03d" $i
$ echo $j
007

--
Best regards | "The only way to really learn scripting is to write
Cyrus | scripts." -- Advanced Bash-Scripting Guide