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?


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.

  • 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 '15 at 6:24
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    @Копать_Шо_я_нашел cron will send an email with the error message, – Jasen Jan 1 '16 at 6:50
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    date +\%Y\ \%m\ \%d\ \%H:\%M:\%S-cronlog – DevilCode Apr 4 '16 at 13:36

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


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


0 * * * * sh jobs.sh

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:


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
  • Output date format will retrun like cron_20180123.log – bala4rtraining Jan 24 '18 at 13:51
  • (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”? – G-Man Dec 23 '18 at 6:27

All of the above answers use double quotes (not all of them worked for my setup). This worked for me:

0 5 * * 3 /data/script.sh > /data/script_`date +\%y\%m\%d`.log 2>&1
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    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.) – G-Man Dec 23 '18 at 6:27
  • The accepted answer simply doesn't work for me. This one does. – Manuel Schmitzberger Dec 23 '18 at 8:44

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