How can I execute `date` inside of a crontab job?

Question

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?

Answer 1

Short answer:

Escape the % as \%:

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

This also uses $(...) instead of the deprecated `...` syntax for command substitution and quotes the expansion of said command substitution.

Long answer:

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

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

This is also confirmed by the second error message you 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.

Answer 2

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:

Answer 3

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

Answer 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

Answer 5

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

Answer 6

This worked for me:

0 5 * * 3 /data/script.sh > /data/script_`date +\%y\%m\%d`.log 2>&1