PHP DateDiff問題

Question

使用下面的PHP代碼，我期望得到'2'作爲我的輸出。但我得到'1'。

有誰知道這是爲什麼？

$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 

非常感謝您的任何幫助。

Answer 1

這麼多的計算...我認爲它的你有一個四捨五入的問題，捨去所有的時間測量...這是你在做什麼更簡潔的外觀：

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

Answer 2

$d1 = new DateTime('2011-03-28'); 
$d2 = new DateTime('2011-03-26'); 

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

輸出：2