PHP计算年龄与出生时间(Y-m-dTH:i:sO)

Question

我试图在php中计算年龄，但只是想增加年龄，如果出生时间也达到，而不仅仅是生日日期。以下是我目前得到的结果:

$today = date("Y-m-dTH:i:sO");
$bday = "1987-01-01T15:30:00+0200";
$diff = abs(strtotime($bday) - strtotime($today));
$age = floor($diff / (365*60*60*24));

如果到达日期，它会工作，但不具有时间敏感性。谢谢你的帮助!

Answer 1

 # object oriented
    $from = new DateTime('1987-01-01 T15:30:00+0200');
    $to   = new DateTime('today');
    echo 'Years '.$from->diff($to)->y; echo '<br/>';
    echo 'Month '.$from->diff($to)->m; echo '<br/>';
    echo 'Days '.$from->diff($to)->d; echo '<br/>';
    echo 'Hours '.$from->diff($to)->h; echo '<br/>';
    # procedural
    echo date_diff(date_create('1987-01-01 T15:30:00+0200'), date_create('today'))->y;

<?php
//Convert to date
$datestr="2015-06-11 19:10:18";//Your date
$date=strtotime($datestr);//Converted to a PHP date (a second count)
//Calculate difference
$diff=$date-time();//time returns current time in seconds
$days=floor($diff/(60*60*24));//seconds/minute*minutes/hour*hours/day)
$hours=round(($diff-$days*60*60*24)/(60*60));
//Report
echo "$days days $hours hours remain<br />";
?>