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 =/

score 35 · Accepted Answer · edited Aug 12 '21 at 19:20

35

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

Where 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.)

edited Aug 12 '21 at 19:20

Pablo A

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answered Dec 11 '11 at 01:30

cyberx86

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thanks for the great feedback +1, i tried to add it to the crontab but doing that didnt work =/ it never renders DATEVAR – qodeninja Dec 11 '11 at 01:50
I just did this on a RHEL/CentOS6 compatible system and it gave the expected result (i.e. added a new line to the file with the date). Which option (of the 3 above) did you try, and what was the result - be specific - and was there an error in /var/log/cron? (if you use the script option, a) don't forget to chmod +x and b) try it on its own first (i.e. not through cron)) – cyberx86 Dec 11 '11 at 01:53
1

option 3 =], I had tried to do that before but it didnt render. But now im noticing you did something a lil dif. why did you double up the $($(DATE))? see my update above – qodeninja Dec 11 '11 at 01:57
2

You have a typo in the update you added - it is `$($DATEVAR)` not `$($(DATEVAR))`. Think of it this way - `$()` - that is, the outer bracket - launches a new shell to process whatever is in the brackets. In this case, the variable $DATEVAR is set to a string and is passed to the shell. `$(DATEVAR)` is nothing - because `DATEVAR` is nothing (i.e. is it missing a $ to identify it as a variable). You could also use backticks around `$DATEVAR` to get the same result – cyberx86 Dec 11 '11 at 02:01
fair enough. updated it. lets try this bad boy! -- Hazaaah! it worked lol – qodeninja Dec 11 '11 at 02:02

score 1 · Answer 2 · edited Jul 09 '20 at 12:29

The "problem" is that cron uses the percent sign as a special character. You have to quote it so it is ignored by cron, but not by the shell. Like most of the folks who I saw that spent hours upon hours tinkering around with when and how and where to quote what, I went through that hack-and-slash as well.

Here's the solution I found that has worked on all flavors of *nix:

# Define the percent sign in a variable.
P=%
#
# Now use it in a $() subshell. Don't get freaked about the double quotes
# inside the $(), they are processed by the subshell.
57 2 * * * root /bin/echo "[CRON] at $(date "+${P}d-${P}b${P}Y")" 1>>/tmp/crontab.log 2>&1

As the old saying in the world of Perl: there is always more than one way to do it. I am not saying this is the only way or the best way -- what I am saying is that this works for crontab entries running on CentOS, Debian, Ubuntu, Mac, and AIX, so this is what am sticking with.

score 0 · Answer 3 · answered Feb 15 '21 at 17:03

Just a working example of using variables in the crontab file and their substitution in the strings:

CURRENT_TIME=date +%Y.%m.%d_%H:%M:%S.%3N
CURRENT_DATE=date +%Y_%m_%d

SIMPLE_VAR=the_simple_var
LOG_DIR=/var/log/cron

* * * * * /bin/echo "simple variable test! ${SIMPLE_VAR}__test!" >> "${LOG_DIR}/test.log"
* * * * * /bin/echo "complex variable test! $(${CURRENT_TIME})__test!" >> "${LOG_DIR}/test.log"

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

3 Answers3

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