removing # from crontab (bash script)
 

Hi All,

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 :-

*/10 * * * * ./aa.sh
* * * * * ./bb.sh
* * * * * ./cc.sh

Please help me.
Thanks in advance
Inder

Check the sed command. Run the file through sed, and remove the "#", such as "sed 's/\#//g'"

Read man sed, and add something like sed s/^[[:space:]]*#// where you want to un-comment a line.

Note that sed normally takes a file as input, so something like sed s/^[[:space:]]*#// /etc/crontab > /etc/crontab.new might work.

You might also think about removing empty lines and trailing comments.

Many Thanks for your help

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.