Shell Programming and Scripting

Hi
I am trying to delete some specific files ( files other than created today) from the server on a cron basis. I wrote a small script, but I am stuck up in how to delete only specific files.


#!/usr/bin/expect -f

set IP [lindex $argv 0]
set timeout -1
spawn ftp $IP
expect ):
send "username\n"
expect word:
send "password\n"
expect ftp>
send "prompt\n"
expect ftp>
send "passive\n"
expect ftp>
send "cd logs/\n"
expect ftp>


After this I am stuck up. It is because I want to delete files with old dates ie, I don't want to delete the file dated today.


The files are like logfile-070501,logfile-070502 and so on..

I cannot run the command "rm logfile-0705*" ( as it will delete today's file too)

By the way, I have the list of files to be deleted in a separate file. I am wondering if by running any other additional script, we can delete selected files from the remote server .

I tried with ! command of ftp, but it runs the commands on local machine.

Your help will be highly appreciated.Thanks

sangfroid,
See if this would work for you:

Code:

vYYMMDD=`date '+%y%m%d'`
vFName='logfile-'$vYYMMDD
rm -f `ls -1 log* | egrep -v "$vFName"`

i cannot execute rm -f inside ftp session and neither can i run egrep in it.

Another option is to prepare a file with all the deletes, adding "delete"
in front of each file name in your list:

Code:

sed 's/^/delete /' list_file_names > to_del_file_names

Would this work for you?

I am getting you. But the problem is how to execute at the remote end "inside" ftp session...

i am stuck up in this one

sangfroid,
Here is a solution:

Code:

rename logfile-070503 hold_logfile-070503
delete logfile*
rename hold_logfile-070503 logfile-070503

you could use co-process something like this:

Click Here