Optimise a shell script / command to delete files

Question

I'm running Magento and it doesn't clean up old session data, so I need to clean it up with a shell script / cronjob:

0,30 * * * * /usr/bin/find /var/www/magento/var/session -name 'sess_*' -type f -mtime +1 -exec rm {} \;

But that script takes usually some minutes to delete the old files. I'm wondering whether I need to search for the files since in this directory are only files that start with "sess_" - but I still need to traverse those files somehow, right?

(around 50.000 files per day are created in this directory)

Answer 1

Don't bother exec'ing rm at all, find can handle it:

0,30 * * * * /usr/bin/find /var/www/magento/var/session -name 'sess_*' -type f -mtime +1 -delete

Answer 2

If the only thing in that directory is sess_* files, you can just leave out the -name 'sess_*', and find will traverse through anyway. Test what it would match by removing the -exec part.

$ /usr/bin/find /var/www/magento/var/session -type f -mtime +1

Then for the cron job,

0,30 * * * * /usr/bin/find /var/www/magento/var/session -type f -mtime +1 -exec rm {} \;

Having said that, I'm not sure that this is where the bottleneck is, so perhaps mikeserv's comments would be more helpful.

Also, you should probably use -execdir instead of -exec for security reasons; see man find.

Answer 3

I would use perl, it is faster and more efficient for this than find + rm

0,30 * * * * cd /var/www/magento/var/session && /usr/bin/perl -e 'for(<sess_*>){((stat)[9]<(unlink))}'

Deleting my log directory with over 500,000 logs in it daily clears it out in under ~6 minutes.