Re: Date addition-reg

2007-11-12 Thread jeevs
On Oct 23, 12:15 pm, [EMAIL PROTECTED] (Jeevs) wrote: > On Oct 23, 2:19 pm, [EMAIL PROTECTED] wrote: > > > > > Hi > > > Thaks for ur valuable info.. > > > my code is > > > my $total_nights="3"; > > $total_nights = "0".$total_nights if ((!($total_nights =~ > > /^0/)) && ($total_nights < 10))

Re: Date addition-reg

2007-10-23 Thread jeevs
On Oct 23, 2:19 pm, [EMAIL PROTECTED] wrote: > Hi > > Thaks for ur valuable info.. > > my code is > > my $total_nights="3"; > $total_nights = "0".$total_nights if ((!($total_nights =~ > /^0/)) && ($total_nights < 10)); > my $start_date="2008-03-24"; > my ($year,$mon,$day) = split/-/,$start_

Re: Date addition-reg

2007-10-23 Thread Jeff Pang
Please don't open a new topic when you repeat the same question. >$day=($day+$total_nights); I've said clearly,you can't do date oprations by this way. See my before message that answered your question fully. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional co

Re: date addition-reg

2007-10-22 Thread Aruna Goke
Jeff Pang wrote: On 10/22/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote: HI I need to get the $to as 18-10-2007 But it throws as 18.can u give the suggestion to get the same format? Thanks #!/usr/perl/bin $from="15-10-2007"; $to=$from+"3"; print "$from,\n"; print "$to"; It looks

Re: date addition-reg

2007-10-22 Thread Jeff Pang
On 10/22/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote: > > > HI > I need to get the $to as 18-10-2007 > But it throws as 18.can u give the suggestion to get the > same format? > Thanks > > #!/usr/perl/bin > $from="15-10-2007"; > $to=$from+"3"; > print "$from,\n"; > print "$to"; > It lo