How do you set a DATE variable to use in a log for crontab output?

Question

I was playing aroudn with some variations of date like

DATE = $(date)

but that didnt work either

crontab -e

CRONLOG=/tmp/log/crontab.log
DATEVAR=`date +20\%y\%m\%d_\%H\%M\%S`
* * * * * echo $DATEVAR >> /tmp/log/crontab.log
*/2 * * * * echo "$DATEVAR hello" >> ${CRONLOG}
*/1 * * * * echo 'every minute' >> ${CRONLOG}

this just outputs the text as is...

I want to create a log entry in crontab.log with a timestamp on each update

How can I do this on CentOS 6?

UPDATE

DATEVAR=date +20%y%m%d_%H%M%S
*/1 * * * * /bin/echo [CRON] $($(DATEVAR)) >> /tmp/log/crontab.log

rendered only [CRON] and NOTHING when I tried it =/

Answer 1

Cron is not a shell - it does not parse commands in the same way that a shell does. As such, your variable is assigned as if it was static text.

There are three solutions I know of to this problem:

Option 1: Use a shell script to generate your command, include whatever variables and logic you want - and call that shell script from cron.

* * * * * /path/to/myscript.sh

myscript.sh:

DATEVAR=`date +20\%y\%m\%d_\%H\%M\%S`
echo $DATEVAR >> /tmp/crontab.log

Option 2: Include the date command directly in your command, and, since the entire command is passed to the shell, the date will be processed and replaced with an actual date.

* * * * * /bin/echo `date +20\%y\%m\%d_\%H\%M\%S` >> /tmp/crontab.log

Option 3: Set the string variable in cron, and pass that to your command to be processed (note - the percent signs do not need to be escaped, and the variable itself is wrapped in $() to execute it in a separate shell - backticks should work the same):

DATEVAR=date +20%y%m%d_%H%M%S
* * * * * /bin/echo $($DATEVAR) >> /tmp/crontab.log

(In all the cases above, you can, of course, use a variable for the log path, instead of 'hard coding' it.)