removing # in crontab (bash script)

[inderPunj]

Hi All,

I am new to scripting. i am creating a script by which it will check all the entries in crontab and print all the enteries start with # and after that it 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 first it print "ok" if # is present at the beginning of the line and then
I want to remove #
And please let me know how can i remove # in crontab through a script. so that the crontab -l should like this :-

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

Please help me.
Thanks in advance
Inder

[Cabhan]

I'm afraid I don't entirely understand.

If we imagine that crontab -l looks like:

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

You want the output to be:

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

?

In any case, checking if a line starts with a '#' can be easily accomplished with sed if you want to use it, or grep if you want to just check within the script. The regular expression is "^#", where '^' indicates the beginning of the string, and the '#' is a literal '#'.