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cron problem!
The cron daemon on one of my HP boxes is giving the following error:
Quote:
Quote:
The /var/adm/cron/queuedefs file does not have a line for cron, so it should be picking up the default value. But in this post, Perderabo says that the default value is also 100. If that is the case, have I already reached the maximum? And if so, do I have any choice, other than having a cronjob that restarts cron every day or so? -EDIT: Here's the uname -a o/p: HP-UX xxxxx B.11.11 U 9000/800 |
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Look up the pid of the cron process. You will have 100 processes with that number as the ppid. This not a matter of a clean exit. They are still running.
Or maybe you have too much stuff in another queue? Notice the language on "man queuedefs": "the total number of jobs that can be run on all the queues is limited to 100". I have never really known whether or not to believe that sentence. |
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Thanks for the inputs. We did move a few cronjobs around, and I added a line in the queuedefs file: c.100j2n60w. One question though: does a script count as one 'job'? So if I have 5 find commands running through cron and I put them in a single script, I will have reduced the number of jobs by 4 right?
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cron is waiting for its child process to finish. It doesn't matter what that child does including running other processes. Want to cheat? Write a cron script that launches a second script in the background and immediately exits.
#! /usr/bin/ksh /path/to/real/script & exit 0 I have never tried this, but I'm guessing it will fake cron out. |
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