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I have a cron job that copies its log file daily to my home folder.

Everyday it overrides the existing file in the destination folder, which is expected. I want to preserve the log from previous dates so that next time it copies the file to destination folder, it preserves the files from previous dates.

How do I do that?

codeforester
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persist
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    What's your command in crontab? I think you could add a date to your log file name, such as `filename_20170120.log` to avoid override. – zhm Jan 20 '17 at 03:28
  • Possible duplicate of [Append current date to the filename via Cron?](https://stackoverflow.com/q/9110663/608639), [Sending cron output to a file with a timestamp in its name](https://serverfault.com/q/117360), [How to add the logs to a crontab with time stamp](https://askubuntu.com/q/391542), [How can cron output to a new log file based on date?](https://stackoverflow.com/q/27539083/608639), [How to log cron jobs?](https://stackoverflow.com/q/4811738/608639), etc. – jww Oct 30 '19 at 21:54
  • why not using logrotate for it? – hariszaman Apr 07 '21 at 11:58

2 Answers2

4

The best way to manage cron logs is to have a wrapper around each job. The wrapper could do these things, at the minimum:

  • initialize environment
  • redirect stdout and stderr to log
  • run the job
  • perform checks to see if job succeeded or not
  • send notifications if necessary
  • clean up logs

Here is a bare bones version of a cron wrapper:

#!/bin/bash

log_dir=/tmp/cron_logs/$(date +'%Y%m%d')
mkdir -p "$log_dir" || { echo "Can't create log directory '$log_dir'"; exit 1; }

#
# we write to the same log each time
# this can be enhanced as per needs: one log per execution, one log per job per execution etc.
#
log_file=$log_dir/cron.log

#
# hitherto, both stdout and stderr end up in the log file
#
exec 2>&1 1>>"$log_file"

#
# Run the environment setup that is shared across all jobs.
# This can set up things like PATH etc. 
#
# Note: it is not a good practice to source in .profile or .bashrc here
#
source /path/to/setup_env.sh

#
# run the job
#
echo "$(date): starting cron, command=[$*]"
"$@"
echo "$(date): cron ended, exit code is $?"

Your cron command line would look like:

/path/to/cron_wrapper command ...

Once this is in place, we can have another job called cron_log_cleaner which can remove older logs. It's not a bad idea to call the log cleaner from the cron wrapper itself, at the end.

An example:

# run the cron job from command line
cron_wrapper 'echo step 1; sleep 5; echo step 2; sleep 10'

# inspect the log
cat /tmp/cron_logs/20170120/cron.log

The log would contain this after running the wrapped cron job:

Fri Jan 20 04:35:10 UTC 2017: starting cron, command=[echo step 1; sleep 5; echo step 2; sleep 10]
step 1
step 2
Fri Jan 20 04:35:25 UTC 2017: cron ended, exit code is 0
codeforester
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  • A wrapper can be helpful but note that cron's loglevel 7 already logs start, end, the command and its error + exit code. Cf. man(8) (referring to Vixie Cron, of course). – xebeche Aug 10 '20 at 09:20
0

Insert 

`date +%F`

to your cp command, like this:

cp /path/src_file /path/dst_file_`date +%F`

so it will copy src_file to dst_file_2017-01-20

upd:

As @tripleee noticed, % character should be escaped in cron, so your cron job will look like this:

0 3 * * * cp /path/src_file /path/dst_file_`date +\%F`
pumbo
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    Notice though that you can't do this in `crontab` because `cron` uses `%` to signify a newline. In your `cron`, you will need to backslash-escape any literal `%`. – tripleee Apr 18 '19 at 09:26