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another DateDIff question

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Jul 27.2006

Hi

My code works but retuns an unformated value

age returns a value of 728559

Birth date was April 17 2001

how do I convert that to an age

[code=php]
<?
function dateDiff($dformat, $endDate, $beginDate)
{
$date_parts1=explode($dformat, $beginDate);
$date_parts2=explode($dformat, $endDate);
$start_date=gregoriantojd($date_parts1[0], $date_parts1[1], $date_parts1[2]);
$end_date=gregoriantojd($date_parts2[0], $date_parts2[1], $date_parts2[2]);
return $end_date – $start_date;
}

?>
[/code]

[code=php]
$age = dateDiff(“/”, $dob,date(‘m/d/y’));
[/code]

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PHP

1 Comments(s) ↴

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@bokehJul 27.2006 — Your two major problem are: [list=1]
  • [*]not using a capital "Y" in your call to the date function (lower case "y" returns "06" which corresponds to 6AD)

  • [*]arguments 2 and 3 when calling your user defined function are in the wrong order

    • [/list]    You also need to take leap years into consideration. Maybe something like this: [code=php]function age($dob /* format m/d/Y */, $delimiter ='/')
    {
    if(!function_exists('is_leap'))
    {
    function is_leap($year)
    {
    return (($year%4)?false:(($year%400)?true:false));
    }
    }
    list($month, $day, $year) = explode($delimiter, $dob);
    $days = gregoriantojd($m = date('m'), date('d'), $Y = date('Y'))
    - gregoriantojd($month, $day, $year);
    $age = 0;
    while($days >= ($remove = (is_leap((($m < 3))?--$Y:$Y--) ? 366 : 365)))
    {
    $days -= $remove;
    $age++;
    }
    return $age;
    }

    // test it
    echo age('04/17/2001');[/code]
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