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I want to create a log file for a cron script that has the current hour in the log file name. This is the command I tried to use:

0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log

Unfortunately I get this message when that runs:

/bin/sh: -c: line 0: unexpected EOF while looking for matching ``'
/bin/sh: -c: line 1: syntax error: unexpected end of file

I have tried escaping the date part in various ways, but without much luck. Is it possible to make this happen in-line in a crontab file or do I need to create a shell script to do this?

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  • 3
    I wasted an hour of my life before googling for this! It never occurred to me that anything needed escaping. Nov 24, 2019 at 6:41

6 Answers 6

268

Short answer:

Escape the % as \%:

0 * * * * echo hello >> ~/cron-logs/hourly/test`date "+\%d"`.log

Long answer:

The error message suggests that the shell which executes your command doesn't see the second back tick character:

/bin/sh: -c: line 0: unexpected EOF while looking for matching '`'

This is also confirmed by the second error message your received when you tried one of the other answers:

/bin/sh: -c: line 0: unexpected EOF while looking for matching ')'

The crontab manpage confirms that the command is read only up to the first unescaped % sign:

The "sixth" field (the rest of the line) specifies the command to be run. The entire command portion of the line, up to a newline or % character, will be executed by /bin/sh or by the shell specified in the SHELL variable of the cronfile. Percent-signs (%) in the command, unless escaped with backslash (\), will be changed into newline characters, and all data after the first % will be sent to the command as standard input.

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    Sorry for my ignorance, but where do you see this error messages? When I do 'grep CRON /var/log/syslog' I see no error messages, although cron failed - kagda.ru/i/9a016249a39_20-05-2015-09:22:47_9a01.png
    – Tebe
    May 20, 2015 at 6:24
  • 3
    @Копать_Шо_я_нашел cron will send an email with the error message,
    – Jasen
    Jan 1, 2016 at 6:50
  • 5
    date +\%Y\ \%m\ \%d\ \%H:\%M:\%S-cronlog
    – DevilCode
    Apr 4, 2016 at 13:36
  • 1
    @Tebe he sees them in the question. if not clear better be comment to the question. My guess that errors are seen in that file because the job is echo hello >> file Dec 5, 2021 at 2:33
16

If you would like to make the date formatting string as a variable (to avoid duplicating the whole string), DO NOT escape % and DO NOT put it in $()

For example, while declare the string, just write:

DATEVAR=date +%Y%m%d_%H%M%S

Then, write cron statement with $($VARIABLE_NAME) like this:

* * * * * /bin/echo $($DATEVAR) >> /tmp/crontab.log

Thanks to cyberx86, her/his answer at ServerFault might be more completed:

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  • 6
    DATEVAR="date +%Y%m%d_%H%M%S"
    – Frank Fang
    Aug 3, 2019 at 8:45
  • This answer has issues with quoting, and it's unclear where to set the DATEVAR variable and how to do it in such a way that it support e.g. format string containing spaces.
    – Kusalananda
    Jan 11, 2021 at 9:47
8

You can also put your commands into a shell file and then execute the shell file with cron.

jobs.sh

echo hello >> ~/cron-logs/hourly/test`date "+%d"`.log

cron

0 * * * * sh jobs.sh
4

In cron, you can use this simple syntax:

*/15 01-09 * * * sh /script.sh >> /home/username/cron_$(date -d"-0 days" +\%Y\%m\%d).log 2>&1
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    Output date format will retrun like cron_20180123.log Jan 24, 2018 at 13:51
  • 6
    (1) What are you saying that hasn’t already been said by the accepted answer?   (2) Your answer is much more complicated than the question.  For example, you added the -d option, which is not used in the question (and you did not explain it).  How do you justify calling this “simple syntax”? Dec 23, 2018 at 6:27
2

A basic solution:

  • use $() for executing date command and return output
  • format datetime to UTC, escape the % character with \
  • add 2>&1 at the end for streaming both stdout and stderr into that log file

Example:

* * * * * echo "Test crontab log" > /tmp/crontab.log.$(date --utc +\%Y\%m\%d_\%H\%M\%SZ) 2>&1

Output:

ls -lh /tmp | grep log

-rw-rw-r-- 1 ubuntu  ubuntu    17 May  4 05:06 crontab.log.20190504_050601Z
-rw-rw-r-- 1 ubuntu  ubuntu    17 May  4 05:07 crontab.log.20190504_050701Z
1

This worked for me:

0 5 * * 3 /data/script.sh > /data/script_`date +\%y\%m\%d`.log 2>&1
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  • 2
    What are you saying that hasn’t already been said by the accepted answer? Are you saying that it works better without quotes than it does with quotes? (Hint: that’s very unlikely.) Dec 23, 2018 at 6:27
  • The accepted answer simply doesn't work for me. This one does. Dec 23, 2018 at 8:44

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