RE: How to use sed to remove a line from a cron file and also remove all commented lines directly above it

In the text below, I want to remove every line from the line beginning with Purge 

through the line containing the pattern $BACKUPDIR/runBackup.sh . 

Any ideas? 



55 10 * * * /opt/company/runThisCommand.sh 

# Purge tables and rebuild database indexes # The first five crontab fields are integer patterns that specify the following: 
#     minute (0-59),
#     hour (0-23),
#     day of the month (1-31),
#     month of the year (1-12),
#     day of the week (0-6 with 0=Sunday). 
45 22 * * * . /etc/profile; . $BACKUPDIR/runBackup.sh 

0 1 * * * /opt/company/runAnotherCmd.sh
====================================================================
I realize I'm cheating here but 
crontab -l | grep -p -v 'Purge tables' > new.crontab

will do the trick.  The '-p' flag treats blocks of text separated by blank lines as a single unit.  The '-v' flag excludes that block of text.
After verifying new.crontab, move it into place.

I realize I'm suggesting unix commands in place of sed and don't intend to offend this group.

Jim Schneider

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