i've created a script which should run every last day of the month. what would be the exact crontab entry for this? thanks!

Have cron run a script on days 28, 29, 30 and 31 of every month.
Create two variables in the script, one containing today's day of the
month and another containing tomorrow's day of the month:

 TODAY=`date +%d`
 TOMORROW=`date +%d -d "1 day"`

 # See if tomorrow's day is less than today's
 if  [  $TOMORROW  -lt  $TODAY  ];  then
         echo  "This is the last day of the month"
 #       Do stuff...
 fi

to be ran at 11pm on the last day of a month:

00 23 * * *[ `date +%d` -eq `echo \`cal\` | awk '{print $NF}'` ] && myJob.sh

You can try as,

00 23 * * * [[ $(date +'%d') -eq $(cal | awk '!/^$/{ print $NF }' | tail -1) ]] && job 1>/dev/null 2>&1

Just a note - when using commands in cron always use the full path to commands...

/usr/bin/date
/usr/bin/cal

In this case however, the commands should work without the full path because cron sets up a very basic environment that usually includes /usr/bin in PATH, but I always err on the side of caution and provide absolute paths....

</pedant>

Cheers
ZB

Was to turn the date math trick shown above into a script that I named 'last-day-of-month.sh':

#!/bin/bash

TODAY=`/bin/date +%d`
TOMORROW=`/bin/date +%d -d "1 day"`

# See if tomorrow's day is less than today's
if [ $TOMORROW -lt $TODAY ]; then
        exit 0
fi

exit 1

12 0 * * * last-day-of-month.sh && /run/my/cron/job.sh

I had a similar dilemma where I need to run a backup script every 2nd Tuesday of the Month. So I added the following to my script:

TODAY=`/bin/date +%d | cut -d"0" -f2`
TUE=`cal | awk {'print $3'} | xargs | /usr/bin/cut -d" "  -f3`
#See if today matches the 2nd Tuesday of the month
if [ "$TODAY" -ne "$TUE" ] ; then

        exit 0

else