Date Calculations in Shell Scripts

Let's say we are dumping a database into a file for backup purposes. We want to keep 3 days of backups around. We could of course rotate the filenames (similar to what newsyslog does: basename.gz, basename.0.gz, basename.1.gz). But we want to backup the directory to a remote host using rsync - and renaming the files would mean to transmit all files with each remote backup. A better naming scheme is to use names with dates, like basename_20070115.gz. That way names don't change and rsync doesn't retransmit yesterday's file. The problem then is that we have to delete backup files older than 3 days. We need a way to calculate dates in a shell script...

The secret sauce in the (partly pseudo) code is that we use the "unix epoch" - seconds since 1970-1-1 - for our calculations. Let's go through the code:

#!/bin/sh
cd /backup/directory

budate=`date '+%Y%m%d'`
                        # today, normal date format, 
                        # backticks to get the at the string
bufilename=basename_`date '+%Y%m%d'`.gz
                        # same thing rolled into filename

# now delete the file from 3 days back, 
# so we have always 3 days of backups
let deletedate=`date '+%s'`-259200
                        # todays date in seconds - 3 days in seconds
                        # key points are using `let' for sh calculations
                        # and using `date' with the %s format string
delfile=basename_`date -r $deletedate '+%Y%m%d'`.gz
rm $delfile

dosomebackupscript -o $bufilename
                        # create the new file

This works by calculating the filename for the file 3 days ago, and deleting that file. After that is done, a different backup script creates the latest file with the same naming scheme. The script runs in a cron job, when it doesn't run on one day, then a stale backup copy is left over. We're not attempting to solve that here.