UNIX for Dummies Questions & Answers

My apache logs are saved in the following format (using rotatelogs):

Oct 8 01:59 access_log.1002412800
Oct 9 01:55 access_log.1002499200
Oct 10 01:58 access_log.1002585600
Oct 11 01:56 access_log.1002672000
Oct 12 01:59 access_log.1002758400

I would like to run a cronjob once a week to delete the log files older than the current date.

How can I accomplish that?

TIA

jamesbond

The first thing that popped into my mind was to use the <i>find</i> command. Something along the lines of:
<i>find /wherever/apache/logs -name access_log.* -mtime 1 | xargs rm</i>

I like using xargs because I'm lazy and don't like messing with the -exec option, although that would work too... You want to create some dummy files in a tmp directory to try this on, because I'm not 100% sure if I wrote that right.

Check the man page for find for more details / options.

Thank you for your help!

I ended up with this :

find /path_to_apache_logs -name "access_log.*" -mtime 1 -exec rm {} \;

One thing I'm wondering about:

I tried 3 things regarding the filenames:

1. single quotes
-name 'access_log.*' -mtime 1 - exec {} \;

2. double quotes
-name "access_log.*" -mtime 1 - exec {} \;

3. filename included in path
path_to_apache_logs/access_log.* -mtime 1 - exec {} \;

All these 3 methods seem to work, but is there a difference in for example using single or double qoutes?


Once again thank you for your assistance!

JamesBond