Strip first part of filename with awk statement 1

[mrimagepueblo]

I need to write a bash shell script to strip the first part of the filename in a directory of files.
The original filenames are
05-12345_1.jpg
05-12345_2.jpg
05-12345_3.jpg
06-56789_1.jpg
06-56789_2.jpg
06-56789_3.jpg
06-56789_4.jpg
etc.
I simply want to strip out the 05- or 06- whatever happens to be in the front, so the end results are:
12345_1.jpg
12345_2.jpg
12345_3.jpg
56789_1.jpg
56789_2.jpg
56789_3.jpg
56789_4.jpg

An awk statement would be preferred but I'm not really picky at this point.

 

[PHV]

A starting point:
ls /path/to/dir | awk '/^..-.*\.jpg$/{print substr($0,4)}'

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[mrimagepueblo]

That command echos to the console fine but the files are not renamed when it's complete?

 

[PHV]

Where did you say that you wanted to rename the files ?
cd /path/to/dir && ls ??-*.jpg | awk '{print mv" "$1" "substr($1,4)}' | sh

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[mrimagepueblo]

The same directory is fine if that is possible. If not I could move them to a temp directory /path/to/dir/temp.

 

[PHV]

Have you tried my suggestion ?

 

[mrimagepueblo]

I did, I get a command not found.
I was even in the directory and just ran
ls ??-*.jpg | awk '{print mv" "$1" "substr($1,4)}' | sh
and I get the command not found???

 

[PHV]

OOps, sorry for the typo :~/
ls ??-*.jpg | awk '{print "mv "$1" "substr($1,4)}' | sh

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[mrimagepueblo]

Works like a charm thank you. Thanks for sticking with me.