Find Access Time Script

Hi

  Here is an example of using find with the parameters
you want.

To remove all files named a.out or *.o that have not been accessed for a week and that are not mounted using nfs, enter:

find / \( -name a.out -o  -name '*.o' \) -atime +7 ! -fstype nfs
-exec \ rm {} \;

Note:   The number used within the -atime expression is +7. This is
the correct entry if you want the command to act on files not accessed
for more than a week (seven 24-hour periods).

Ok, but what about for 45 minues?

Unfortunatelly now i am at home, where a have no unix to
do some test.
 But i advise you to see man find for the parameters
 of this command.

  I have already used -atime because i needed to delete
a minimum of one day, not minutes as you want.

  Good luck

What about doing this, would it work?

find /directory -mmin +40 > temp.file
rm -r < temp.file

I haven't seen the -mmin option (what is your OS?), but the rm won't work.
If you have that -mmin, then do something like:
find /directory -mmin +40 -exec \ rm {} \;

======
Werner

The OS is HP-UX 10.24

Hi, it seems to be difficult (impossible?) to do something with find with smaller steps than one day, so I have another suggestion:
If you list the content of your directory with ls -lu, you will get the access time instead of modification time in the time column.  Futhermore, if you use ls -trlu the list will be sorted with the most recently accessed files at the bottom.
From here, you can probably make a script that analyzes the time colum and make the desired actions.

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Moondancer
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Since we didn't get any feedback from the asker about the precision needed, I claim that I provided a working solution.

======
Werner

Thank you, Werner, and sorry for the delay.  This was finalized today.  There are just too many in my EE Inbox, all the time.  :)
Moondancer - EE Moderator

Don't feel sorry! My revenge is in CS already :-)

=======
Werner