Explain this short Cron job

IamtheWalrus

Can anyone explain to be what this cron job is trying to achieve?

05,40 15,17,19,21,23 * * * * for f in `ls /home/mydev/my_feeds/`; do [ ! -e /sbcimp/dyn/data/my/uat/feeds/input/$f ] && [ ! -e /sbcimp/dyn/data/my/uat/feeds/processed/$f ] && mv /home/mydev/my_feeds/$f /sbcimp/dyn/data/my/uat/feeds/input/; done

Skrynesaver

for f in `ls /home/mydev/my_feeds/`; do     
#for every file in "/home/mydev/my_feeds/" 
	[ ! -e /sbcimp/dyn/data/my/uat/feeds/input/$f ] \  
# if there doesn't exist a file of the same name in /sbcimp/dyn/data/my/uat/feeds/input/
	&& [ ! -e /sbcimp/dyn/data/my/uat/feeds/processed/$f ] \ 
# nor a file of the same name in /sbcimp/dyn/data/my/uat/feeds/processed/
	&& mv /home/mydev/my_feeds/$f /sbcimp/dyn/data/my/uat/feeds/input/; 
# then move the file from /home/mydev/my_feeds/ to  /sbcimp/dyn/data/my/uat/feeds/input/
done

Returning to the time statement, do this at 05 and 40 past the hours 15,17,19,21,23 every day

IamtheWalrus

Fast, accurate and without fuss. Thank you. I'm very grateful.

One question - does ! -e mean 'does not exist'?

Skrynesaver

yup, so my overly quick reading of that script requires review

smcelhinney

Skrynesaver wrote: »

yup, so my overly quick reading of that script requires review

Yup, looks good to me. Very succinctly explained.