while (<>) {
if ( /^#/ ){
print;
}
}
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Thursday, September 12, 2002 3:12 PM
To: [EMAIL PROTECTED];
[EMAIL PROTECTED];
[EMAIL PROTECTED]
Subject: A regular expression question
I have a cron file which the line looks like this:
00 22 * * * perl -S queue_submit.pl esd-foundry "perl -S host-scrubber.pl"
#build host scrubber
30 22 * * * perl -S queue_submit.pl esd-foundry "perl -S group-scrubber.pl"
#build group scrubber
30 22 * * * perl -S queue_submit.pl esd-foundry "perl -S object-scrubber.pl"
#build object scrubber
I want to get the comment after # of each line(not including "#"), how could
I do this?
Thanks a lot in advance!
Lixin
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