Re: A unix cron job question
- From: "Humberto Rodriguez" <sub@xxxxxxxxxxxxxxx>
- To: <programmingblind@xxxxxxxxxxxxx>
- Date: Fri, 25 Jun 2010 20:42:31 -0400
Probably the easiest way is to use PHP and the seconds from the Unix Epoch
(Jan 1, 1970 GMT). To go back 30 days, you may do:
$now= date("U");
$then= $now - (60*60*24*30);
Then, if you want the ISO-8601 date:
$date= date("c", $now);
For the RFC-2822 date:
$date= date("r", $now);
There are many more formats for getting the date and time anyway you want,
all the Unix choices. Look at the function date() in php.net.
HIH,
Humberto
----- Original Message -----
From: "Lafond, Eileen" <Eileen.Lafond@xxxxxxxxxxx>
To: <programmingblind@xxxxxxxxxxxxx>
Sent: Friday, June 25, 2010 6:29 PM
Subject: A unix cron job question
We run many cron jobs daily that create a log file per cron job. This
creates many log files over time. I think that we have about 33000 files
right now.
I am trying to create a cron job that will run once a month and delete all
the log files that are more than thirty days out.
I am trying now just to put the specification document together and I am
having a difficult time with the details of the code.
I was thinking of using the following:
mydate="`date +"%m%d%y%H%M""
I cannot figure out how to write the code to go thirty days back and when I
get that, I will then create a delete statement to delete the rest of the
files.
Does anyone have any idea as to how I can use the mydate above to figure out
thirty days bac?
Or am I doing this the wrong way and there might be an easier way to do it?
Thanks for any help.
Eileen La Fond
Work Phone: (206) 386-0011
Email: Eileen.lafond@xxxxxxxxxxx
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