function isLeapYear($year){
if(checkdate(2,29,$year)){
return $year.' is a leap year';
}else{
return $year.' is not a leap year';
}
}
On Thu, Jun 18, 2009 at 5:51 AM, suhasini<ammana2@xxxxxxxxxxx> wrote:
>
>
> hii
> ur first question is print even numberss..am i right?
> u can do something like this
> <?php
> $i =2;
> while($i < =10)
> {
> echo "$i";
> $i= $i + 2;
> }
> ?>
>
> Thanks & Regards
> Suhasini
>
> --- On Thu, 18/6/09, ellamae.luyun <ellamae.luyun@xxxxxxxxx> wrote:
>
> From: ellamae.luyun <ellamae.luyun@xxxxxxxxx>
> Subject: php codes for the following pls ..
> To: php-objects@xxxxxxxxxxxxxxx
> Date: Thursday, 18 June, 2009, 5:38 AM
>
> what is the syntax/ code to this problems:
>
> 1) 2, 4, 6, 8, 10
>
> 2) 2006 is not a leap year
>
> 2008 is a leap year
>
> 3) if you input, age is 18 then the output should be "your a debutant"
>
> if you input, age is 17 then the output should be "your a minor"
>
> if you input, age is 25 then the output should be "your an adult"
>
>
>
>
>
>
>
>
>
>
>
> Love Cricket? Check out live scores, photos, video highlights and more.
> Click here http://cricket.yahoo.com
>
> [Non-text portions of this message have been removed]
>
>
