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I am new to scripting. i am creating a script by which it will check all the enteries in crontab and will remove # if any entry will be commented. Please help me to resolve out.
For exp :-
$crontab -l
#*/10 * * * * ./aa.sh
* * * * * ./bb.sh
# * * * * * ./cc.sh
I want to remove # so that the crontab -l should like this :-
I am new to scripting. i am creating a script by which it will check all the enteries in crontab and will remove # if any entry will be commented. Please help me to resolve out.
For exp :-
$crontab -l
#*/10 * * * * ./aa.sh
* * * * * ./bb.sh
# * * * * * ./cc.sh
I want to remove # so that the crontab -l should like this :-
How can i run this command through a script so that it will reflect all the enteries in crontab ?
Thanks
Inder
If you read the man page on SED, or look at any of the Google hits you'd get, you'd see the 'g' at the end of the statement, means to replace all occurrences.
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