Forum Moderators: phranque
I'm having a spot of bother with the cron job shown below.
What it suppose to do is:
1. Unzip the zip files in the given directory
2. Move the zip files into a zip directory
3. Move the log file after it has been unzipped into the logs directory.
The trouble I am having is that I'm not quite how to put this all together I've had a go, I've used the pipe command but I don't think that is the right command to use.
Code:
30 09 * * * /usr/bin/unzip /home/sites/test.domain.com/ex*.zip ¦ mv /home/sites/test.domain.com/ex*.zip /home/sites/test.domain.com/zips ¦ mv /home/sites/test.domain.com/ex*.log /home/sites/test.domain.com/logs
Thanks again.
i.e
in script.sh
#!/bin/ksh
/usr/bin/unzip /home/sites/test.domain.com/ex*.zip
mv /home/sites/test.domain.com/ex*.zip /home/sites/test.domain.com/zips
mv /home/sites/test.domain.com/ex*.log /home/sites/test.domain.com/logs
and then in your cron file
30 09 * * * script.sh
Hope this works
John