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Compute 7 days ago in shell script

Hi, I have a simple backup script trigged by cron, below... it creates a backup file each day named for that day-- i.e. "db_01-03-2010.sql.bz2"

This works perfectly.

I want to ADD to this script so it computes the date for 7 days prior, and then removes that file-- so when "db_01-10-2010.sql.bz" is created, it'll figure out the date 7 days before (01-03-2010) and remove that file.

Obviously this is a hair non-trivial and I can't just subtract 7 from the day value, since the date 7 days before 01-03-2010 is 12-27-2009, etc.

How can I do this in a bash shell script?

thanks,
Eric

#!/bin/bash
mysqldump -u mysql_user -pmysql_pass tablename | bzip2 -c > /my/backups/db_`date +%m-%d-%Y`.sql.bz2

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pnoeric
Asked:
pnoeric
1 Solution
 
vikas_madhusudanaCommented:
you can use find command to find the files 7 days older

find <path> -name *.* -mtime +7
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woolmilkporcCommented:
Hi,

if you have GNU date try this:

OLDDATE=$(date -d "7 days ago" "+%m-%d-%Y")
echo rm db_${OLDDATE}.sql.bz2

Of course you can write it as a one-liner:

echo rm db_$(date -d "7 days ago" "+%m-%d-%Y").sql.bz2

I put echo in front of the actual command, so you can verify if it will do what you desire.
Remove it when you're satisfied with the result.

wmp


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pnoericAuthor Commented:
perfect, thank you.
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