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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'。

有人知道为什么是这样吗?

$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

非常感谢您的帮助。

4

2 回答 2

5

这么多计算...我认为这是您遇到的四舍五入问题,四舍五入所有时间测量...这里是您正在做的更简单的外观:

function dateDiff($start, $end) {
  $start_ts = strtotime($start);
  $end_ts = strtotime($end);
  $diff = $end_ts - $start_ts;
  return round($diff / 86400);
}
于 2011-03-23T19:35:24.183 回答
4
$d1 = new DateTime('2011-03-28');
$d2 = new DateTime('2011-03-26');

echo $d1->diff($d2)->d;

输出:2

于 2011-03-23T19:34:56.883 回答