comment out a cron job as part of a script

Shell Programming and Scripting
1

Greetings,
I am creating a ksh script to automate the installation of a utility on many servers.
As part of this install, I want to check for a job in root's crontab.
If the job exists, I need to comment it out.
I know I will need to copy off the crontab then read it back in, but I am struggling on commenting the job out while maintaining the integrity of the crontab.
The job will have the same path, so this is a constant to check for.
The crontab entry might not reside on the same line for each server.
Any tips would be appreciated.
Thanks

2

You can copy off the crontab, grep for the job name, if found, put a hash at the beginning of the line (you can use sed) and write it back.

Show us the code you have tried so far.

3

I am aware of all those steps. It is the syntax of the sed command that I am struggling with. If I just grep for the command and issue the

sed 's/^/#/'
#0 10 * * 5 /my/cron/job

I only get that single line in my new file. I am not strong with sed and need some help with the [address]. I have tried these and get the same error,

$ crontab -l | sed [/my/cron/job]s/^/#/
Unrecognized command: [/my/cron/job]s/^/#/
$ crontab -l | sed '[/my/cron/job]s/^/#/'
Unrecognized command: [/my/cron/job]s/^/#/
4

Thats why you should start with posting code to save everyone's time. Here goes:

$ cat cron.sample
0 10 * * 5 /my/cron/job
0 10 * * 5 /other/cron/job

$ cat cron.sample |sed '/\/my\/cron\/job/s!^!#!'
#0 10 * * 5 /my/cron/job
0 10 * * 5 /other/cron/job
5

Understood on the code posting. I will keep this in mind for future posts.
This works great. Thanks for the quick response, much appreciated.