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Commenting out a cron entry through a shell script

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# 1  
Old 03-27-2013
Commenting out a cron entry through a shell script
In my cron thare is a line like

24 11 * * * /usr/batch/bin/abc.sh > /usr/batch/log/abc.log 2>&1

along with other entries. I want to comment out this line through a shell script. My local variable 'line'ontains the full entry (i.e. 24 11 * * * /usr/batch/bin/abc.sh > /usr/batch/log/abc.log 2>&1
)

Any suggestion?
# 2  
Old 03-27-2013
List it, comment out that line, store it and feed it back in:
Code:
ct2=$( crontab -l | sed 's/^24 11 . . . .*abc.sh /# &/' )
echo "$ct2" | crontab

It might work as one pipeline, but I think " ... | crontab " clears the crontab before you can list it. You can give it a try, once you have a copy of the crontab in a file!
# 3  
Old 03-27-2013
Thanks. But my variable contains value 24 11 * * * /usr/batch/bin/abc.sh > /usr/batch/log/abc.log 2>&1

i.e. line="24 11 * * * /usr/batch/bin/abc.sh > /usr/batch/log/abc.log 2>&1"

And I have to use line in the command.
# 4  
Old 03-27-2013
Well, sometimes regex is a pain.
  • You can do the substitution in a 'while read line2 ; do case "$line2" in ("$line")' loop, as case's file glob meta are more friendly (but dangerously loose). Maybe skip case in favor of: if [ "$line" = "$line2" ]
  • You can 'fgrep -vx "$line2"" and move it to the end as an append: echo "# $line"
  • You can fgrep each line individually (many fork/exec but no cron file is that long), detectiing a blank output with not blank input.
# 5  
Old 03-28-2013
Thanks. Instead of one line command, I created a loop.
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