PHP日期差异问题
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” 有人知道这是为什么吗PHP日期差异问题,php,datediff,Php,Datediff,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” 有人知道这是为什么吗 $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($d
$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
非常感谢你的帮助
$d1 = new DateTime('2011-03-28');
$d2 = new DateTime('2011-03-26');
echo $d1->diff($d2)->d;
输出:
2
如此多的计算。。。我想这是一个取整的问题,所有的时间测量都取整。。。以下是您正在做的事情的简单介绍:
function dateDiff($start, $end) {
$start_ts = strtotime($start);
$end_ts = strtotime($end);
$diff = $end_ts - $start_ts;
return round($diff / 86400);
}
应该注意的是,这只是PHP5.3+。@evolve,非常正确。我不再提及这一点,因为PHP5.2已经过了后期支持阶段。感谢您的帮助。不幸的是,应用程序的其余部分依赖于PHP4:'(