Shell Programming and Scripting

I have 34 file in a directory that all have different names, however, they do have 1 pattern in commmon. They all have "-10-11-2010" date format in the name. I want to replace the date in the file name with a supplied date or maybe even the system date. I am sure I will be using awk or sed to accomplish this. Any help would be great. Thanks in advance.

P.S.
Any ideas why the .bash_profile will not load or be read at login time? I set one up for a user and they have to issue the source command in order for their .bash_profile to take affect. The default .profile is normal and everyone uses it, in fact not many use .bash_profile or a custom .profile

Thanks!

Hi.

To your first question:

Code:

NEW="12-12-2012" # get this value whichever way

ls *-10-11-2010* | while read FILE; do
  mv "$FILE" "${FILE/-10-11-2010/-$NEW}"
done

Or in sed / xargs (no good if the filenames have spaces):

Code:

NEW="12-12-2012"

ls *-10-11-2010* | sed "p; s/-10-11-2010/-$NEW/" | xargs -n2 mv

Awesome. I will give it a shot. Thanks!!!!!

---------- Post updated at 08:38 PM ---------- Previous update was at 08:31 PM ----------

The only issue is, is that the dates in the filename may all be different. Some may say text-09-20-2008-text or text-10-10-2007-text. I was thinking i could just match for numbers, any numbers in the format text-00-00-0000-text and replace what ever number are text-num-num-num-text to the date of my choice. Are there some gen expressions i could use, or some perl one liners?

Wow! thats crazy. I will try it and let you know. Thanks for the help.

Hi,

Another solution:

Code:

find . -type d | sed 's/\(.*\/.*\)[0-9]\{2\}-[0-9]\{2\}-[0-9]\{4\}\(.*\)/mv & \1$(date +%m-%d-%Y)\2/' | sh -x

Regards,

Code:

find . -type d | sed "s/\(.*\/.*\)[0-9]\{2\}-[0-9]\{2\}-[0-9]\{4\}\(.*\)/mv & \1$(date +%m-%d-%Y)\2/" | sh -x