php Code for Year Count

[talmik]

Hello all

This bit of code is supposed to count the number of years between a creation date and the current date then place that number of years in the database. I seem to be missing something or have done something wrong. Can anyone see where it all went wrong?

Quote

function GetMemberYears($P_CharacterID)
{
    $Creation = SQL_ReadField("Characters",$P_CharacterID,"CreationDate");
    $CreationExpand = explode("/",$Creation); 
    $curMonth = date("m");
    $Years = date("y") - $CreationExpand[2]; 
    if($curMonth<$CreationExpand[1] || ($curMonth==$CreationExpand[1] && date("d")<$CreationExpand[0])) 
            $Years--; 

    return $Years;
}

 

[Barand]

Do it the easy way

$theDate = '10/08/2005';

$dt1 = DateTime::createFromFormat('d/m/Y', $theDate);
$years = $dt1->diff(new dateTime())->y;

echo $years;    //--> 12

 

[requinix]

I sure hope you're not 18 years late to the Y2K party.

[Barand]

I wonder if there will be the same amount of panic when the Unix Epoch ends in the 2030's

[requinix]

Y2K38? Surely everything important is on 64-bit (or higher?) computers by then...

[dodgeitorelse3]

I have 64 bit computer but I failed a y2k38 test returned a date in 1969 lol, guess I'll be finding solution to this. Of course it is using

$date = '2040-02-01';
$format = 'l d F Y H:i';

$mydate1 = strtotime($date);
echo '<p>', date($format, $mydate1), '</p>';

but when I use

$date = '2040-02-01';
$format = 'l j F Y H:i';

$mydate2 = new DateTime($date);
echo '<p>', $mydate2->format($format), '</p>';

I get my expected future date

[requinix]

Your computer may be 64-bit but what about your PHP? The first example will work with 64-bit PHP.