Command not working when run from cron

Question

I can successfully run this command in the terminal:

find /home/www/files/1 -type f -name '*' -mtime +180 -exec rm {} \

However when I add it as a cron job like so:

*/5 * * * * find /home/www/files/1 -type f -name '*' -mtime +180 -exec rm {} \

It doesn't work and the system sends a mail saying find: missing argument to '-exec'

What can I do to make it work as a cron task?

Are you sure the first version is working? You appear to have a shell escape `\\`` but no actual character (AFAIK it should be `\;` - or optionally `+` if your version of `find` supports it) — steeldriver, Jan 14 '18 at 01:02
Yes, the first command works as it's written. I will try adding `/usr/bin/find` now. — Dan P., Jan 14 '18 at 01:03
I don’t suppose you’re using the fish or rc shell? https://unix.stackexchange.com/a/247786 — Jeff Schaller, Jan 14 '18 at 01:12
@GeorgeUdosen cron’s default path should include /usr/bin: https://unix.stackexchange.com/q/401621/117549 — Jeff Schaller, Jan 14 '18 at 01:14
I am not aware of fish or rc shell. Added `usr/bin/find` but I still get `usr/bin/find: missing argument to -exec` — Dan P., Jan 14 '18 at 01:15
Oh! It works, I think the error might come though if there's no file found to delete, but haven't confirmed yet. — Dan P., Jan 14 '18 at 01:16
Two additional notes: what do you think `-name '*'` accomplishes? Also, some versions of find support `-delete`. — dhag, Jan 14 '18 at 01:24
`-name '*'` to delete any file that's older than 180 days. Is it even needed? — Dan P., Jan 14 '18 at 01:32

score 2 · Answer 1 · answered Jan 14 '18 at 01:41

2

You are missing the semi colon at the end of the -exec argument.

answered Jan 14 '18 at 01:41

jmw

1 Answers1