[EMAIL PROTECTED] wrote:
> From: sekhar kavuru [mailto:[EMAIL PROTECTED]
>>
>> I need help with a generic function that can give the value of next
>> day Example : 12/31/2005 => 01/01/2006
>>
>> I want to accomplish this without Cal::Date
Fine, but by using timelocal and localtime together, you can generate the point
in time and then add 1 day ( 86400 ) to the point in time to get the next day.
By doing that, it becomes one use statement plus 3 more lines give or take to
get you the next day. You don't have to go through all the other checks.
Wags ;)
>
> Are you saying you want to do it without using a date module, or
> without using that particular date module?
>
> Let me give it a quick shot (off-the-cuff and completely untested):
>
> ------------------------------------------------
> sub nextDay
> {
> $date = shift;
>
> @month_days = ( undef, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31,
> 30, 31 );
>
> ($m, $d, $y) = $date =~ m#(\d+)/(\d+)/(\d+)#;
>
> # Check for leap year
> if ($y % 4 == 0)
> {
> $month_days[2]++;
> }
>
> $d++;
>
> if ($d > $month_days[$m])
> {
> $d = 1;
> $m++;
> }
>
> if ($m > 12)
> {
> $m = 1;
> $y++;
> }
>
> return "$m/$d/$y";
> }
> ------------------------------------------------
>
> Caveats:
> - This assumes the incoming date is valid
> - There needs to be a few more checks for the leap year, but I'm too
> lazy to look it up right now.
>
> Bowie
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