PHP DateDiff问题

Question

使用下面的PHP代码,我希望得到'2'作为我的输出.但我得到'1'.

有人知道为什么吗？

$returndate = preg_replace('#(\d+)/(\d+)/(\d+)#', '$3-$2-$1', '2011-03-28');
$departdate = preg_replace('#(\d+)/(\d+)/(\d+)#', '$3-$2-$1', '2011-03-26'); 

$diff = abs(strtotime($returndate) - strtotime($departdate));

$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

echo $days; // expecting 2, but get 1

非常感谢任何帮助.

Answer 1

如此多的计算...我假设它有一个舍入问题,所有时间测量四舍五入...这里有一个更简单的看看你在做什么:

function dateDiff($start, $end) {
  $start_ts = strtotime($start);
  $end_ts = strtotime($end);
  $diff = $end_ts - $start_ts;
  return round($diff / 86400);
}