通过那里没有任何东西是有价值的（生日）

Question

if ($_SERVER["REQUEST_METHOD"] == "POST") {

        $email          = $_POST['email'];
        $viewemail      = $_POST['viewemail'];
        $gender         = $_POST['gender'];
        $birthdayyear   = 0 + $_POST['birthdayyear'];
        $birthdaymonth  = 0 + $_POST['birthdaymonth'];
        $birthdayday    = 0 + $_POST['birthdayday'];
        $country        = 0 + $_POST['country'];
        $lang       = 0 + $_POST['lang'];
        $favpet     = $_POST['favpet'];

        if (strlen($birthdayyear) == 2)  {
        $fourdig = date_create_from_format('y', $birthdayyear);
        $birthdayyear = date_format($fourdig, 'Y');
        if ($birthdayyear > date('Y'))
        $birthdayyear = $birthdayyear - 100;
        }


    die($birthdayyear);

the page will not display the value, but if i write die('sd'.$birthdayyear); I'll get "sd1988". This is when I'm posting the value "1988" but for some reason it works if I just post "88".. So what am I'm doing wrong?

Thanks

Answer 1

Added "0 + (int)$_POST[]" was the right solution to this. thanks Shane Henry. Anyway, heres the working code if someone wants it

if ($_SERVER["REQUEST_METHOD"] == "POST") {
$birthdayyear   = 0 + (int)$_POST['birthdayyear'];
$birthdaymonth  = 0 + (int)$_POST['birthdaymonth'];
$birthdayday    = 0 + (int)$_POST['birthdayday'];

    if (strlen($birthdayyear) == 2)  { // if the year value is "88", it will convert it to a four digit number.
        $birthdayyear = date_format(date_create_from_format('y', $birthdayyear), 'Y');
    if ($birthdayyear > date('Y')) // if the year is over the present year
        $birthdayyear = $birthdayyear - 100; // 2035 will become 1935
    }

$birthday = date("Ymd", strtotime($birthdayyear.'-'.$birthdaymonth.'-'.$birthdayday)); // returns "YYYYmmdd"
echo $birthday;
}

Thanks for all the help guys!