guys,

I have a need for a script that will delete all files in a given directory that are over 2 hours old. I will set this up to run in cron. I'm having a little trouble coming up with the syntax that will identify these files. Is there a variation of the ls command that I can use to identify these files? Any help will be appreciated.

Thanks!

Use the "date" command to get the current date and time. Then subtract 2 from the hours. This may require adjusting the date if you are close to midnight. Once you have the date and time of 2 hours ago, use the "touch" command to create a temp file with that mtime. Next, use the "find" command is get a list of files older than the temp file and delete them. Finally, remove that temp file.

mmm tough one... I know how to do this for files that were modified 'n' number of days ago...but I'm not sure about hours....

find . -mtime +5 -exec rm {} \;

This will remove files over 5 days old..... but the sytax for mtime states days only.... i tried decimals..but no joy.... anyone know if mtime or an equivalent can be used for hours?

I know the date command when formatted with give you the time.

Example:

phdb102 [ogd] >date +%H:%M
12:02

This is the using variables in a shell script:

#!/usr/bin/ksh

the_hour=`date +"%H"`
the_min=`date +"%M"`
calc_time=`expr $the_hour - 2`
echo "The current time is $the_hour:$the_min"
echo "This is the calculated time $calc_time:$the_min"

The result:

phdb102 [ogd] >the_time.sh
The current time is 12:04
This is the calculated time 10:04

Knowing this I am not sure how to incorporate this into the find command. I will experiment and let you know what I come up with.