Cron fails on single apostrophe

Question

The following does work as expected:

date +'%d-%b-%Y-%H-%M'

28-Sep-2009-14-28

But none of the following 4 entries from crontab are working.

* * * * * date +\'%d-%b-%Y-%H-%M\' >> /backup/shantanu/testing.txt
* * * * * date +'%d-%b-%Y-%H-%M' >> /backup/shantanu/testing1.txt
* * * * * date +"%d-%b-%Y-%H-%M" >> /backup/shantanu/testing2.txt
* * * * * date +\"%d-%b-%Y-%H-%M\" >> /backup/shantanu/testing3.txt

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

I can save the same code in a shell script and set the cron, but I will like to know if it is possible to directly set a cron for the task.

The actual cron entry that I am trying to set looks something like this...

16 * * * * mysqldump myDB myTB > /backup/ABCbc$(date +'%d-%b-%Y-%H-%M').sql 2> /backup/ABCbc_errORS$(date +'%d-%b-%Y-%H-%M').txt

Answer 1

There are four common causes for cron job commands to behave differently compared to commands typed directly into an interactive shell:

• Cron provides a limited environment, e.g., a minimal $PATH, and other expected variables missing.
• Cron invokes /bin/sh by default, whereas you may be using some other shell interactively.
• Cron treats the % character specially (it is turned into a newline in the command).
• The command may behave differently because it doesn't have a terminal available.

You must precede all % characters with a \ in a crontab file, which tells cron to just put a % in the command, e.g.

16 * * * * mysqldump myDB myTB > "/backup/ABCbc$(date +'\%d-\%b-\%Y-\%H-\%M').sql" 2> "/backup/ABCbc_errORS$(date +'\%d-\%b-\%Y-\%H-\%M').txt"

(As a separate matter, always put double quotes around a "$variable_substitution" or a "$(command substitution)", unless you know why not do it in a particular case. Otherwise, if the variable contents or command output contains whitespace or ?*\[, they will be interpreted by the shell.)