[SOLVED] /usr/bin/run-parts from current seems buggy

Thom1b · 01-17-2024, 04:29 AM

Quote:

Originally Posted by Windu

You are confusing '|' (the pipe symbol) with '||' (the logical OR). The script pipes the output of the "$i" command into the 'sed'. That executes every time, not only when the previous command exits with a success code.

With « $i | sed '1i\ », the name of the job is displayed only if $i return 0, I don't understand why, but it works with « $i 2>&1 | sed '1i\ »
I made many tests before posting my message, I'm 99% sure I'm right. Maybe you can test and confirm if I'm wrong or right?

Thom1b · 01-18-2024, 09:52 AM

Hi,

The issue with sed is it does nothing if stdin is empty.

I think I found a not too bad fix, still not perfect. I replaced the lines 92 to 94 in /usr/bin/run-parts:

Code:

$i | sed '1i\
'"$i"':\
'

With these:

Code:

if result=$($i 2>&1); then
  [[ -n "$result" ]] && echo -e "$i finished successfully:\n\n$result"
else
  echo -e "$i finished with error(s):\n\n$result" 1>&2
fi

With this crontab:

Code:

47 * * * * /usr/bin/run-parts /etc/cron.hourly

stdout and stderr are sent by mail as expected with the name of the jobs and report if the job is done successfully or with error.

With this crontab:

Code:

47 * * * * /usr/bin/run-parts /etc/cron.hourly 1> /dev/null

a mail is sent only if a job failed but stdout and stderr are both sent.
Right now I didn't find a good way to send only stderr, but it's a start.

Windu · 01-18-2024, 10:28 AM

Quote:

Originally Posted by Thom1b

With « $i | sed '1i\ », the name of the job is displayed only if $i return 0, I don't understand why, but it works with « $i 2>&1 | sed '1i\ »
I made many tests before posting my message, I'm 99% sure I'm right. Maybe you can test and confirm if I'm wrong or right?

Well yes, if the output from "$1" is empty then the sed will still be executed but it will not output anything (the empty input results in empty output) and cron will not invoke the mailer.
If the "$1" command writes text to STDOUT, then the "sed" command will simply add a header line containing the name of the program and ending with a colon, then an empty line; and then the actual output. And that total gets emailed to you so that you know what program this is from.

Thom1b · 01-18-2024, 10:30 AM

I think about always redirecting run-parts to 1>/dev/null. Instead of, maybe we can use the /etc/default/run-parts file and specify something like "LOGLEVEL=${LOGLEVEL:-err}". I can work on it if you're interested.

Thom1b · 01-18-2024, 10:35 AM

Quote:

Originally Posted by Windu

Well yes, if the output from "$1" is empty then the sed will still be executed but it will not output anything (the empty input results in empty output) and cron will not invoke the mailer.
If the "$1" command writes text to STDOUT, then the "sed" command will simply add a header line containing the name of the program and ending with a colon, then an empty line; and then the actual output. And that total gets emailed to you so that you know what program this is from.

You know what program this is from when STDOUT is not empty, you're right. If STDERR is not empty but STDOUT is, sed won't add the name of the program. That's the issue.