Shell Programming and Scripting

Okay, here is what I have:

Code:

#!/bin/sh
cd /opt/backup/app || exit
find . -name "*.tar" -mtime +7 -exec gzip {} \;
find . -name "*.war" -mtime +7 -exec gzip {} \;
find . -name "*.tar.back*" -mtime +7 -exec gzip {} \;
#find . -name "*.gz" -type f -mtime +91 -exec rm {} \;

Here is a little bit on what I want to do. The /opt/backup/app directory has many sub-directories with many more directories under them. Now, everything in these directories are backups of old files, in this case all the files I am looking to execute on are *.tar, *.war. and *.tar.back*

What I am doing now, is just gzipping anything older than 7 days, and deleting anything older than 91 days. Now, I ran into a situation where it would have been helpful to have one of these files after 91 days. So I was wondering how I would change this script to the following:

Run the find command(or some equivelant if it's easier that way), then a logic statement to say, if this .gz file is the only one in this directory, then do nothing, else delete all .gz files older than 91 days, but keep the most recently updated file.

lol, well I was going to write it in some kind of code-block, but then I didn't.

Okay, so worked exactly as I had explained, unfortunately I didn't explain it correctly.

Rather than if this it he only .gz in a directory, do not delete, I meant to say, to always leave 1 .gz file in the directory. Regardless of how old it is. See, these are backup files that we may need, so we want to keep at least one file for quickly reverting changes if needed. So, let's say in one directory I have 8 files, and they are older than the 91 days, I want the script to delete the 7 oldest .gz files, and keep the one that's been modified the most recently.