PHP比較兩個日期

Question

林之間的差異試圖返回兩個日期之間的區別，我根據example found on stackoverflow

我的問題的工作？我得到完全錯誤的結果返回，下面的代碼返回30年，0個月，9天，當它顯然應該只有7天或1周。下面

代碼如下：

date_default_timezone_set('America/Los_Angeles'); 

    $pickupDate = '2016-10-13'; 
    $returnDate = 2016-10-20; 

    $diff = abs(strtotime($pickupDate) - strtotime($returnDate)); 
    $years = floor($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)); 

    printf("%d years, %d months, %d days\n", $years, $months, $days); 

任何輸入讚賞

Answer 1

只要把單引號中的歸期像$returnDate = '2016-10-20';，你可以使用PHP像date_diff()功能，

$daysdiffernce = date_diff(date_create('2016-10-13'),date_create('2016-10-20')); 
echo $daysdiffernce->format("%R%a days"); 

，這會給完全+ 7天回答

Answer 2

首先，代碼不考慮閏年，變個月之類的東西的長度。

實際上有一個PHP函數對於這一點，請檢查鏈接的詳細信息：採取http://php.net/manual/en/datetime.diff.php，並舉例：

$datetime1 = new DateTime('2016-10-13'); 
$datetime2 = new DateTime('2016-10-20'); 
$interval = $datetime1->diff($datetime2); 
echo $interval->format('%y years, %m months, %d days'); 

Answer 3

試試這個，它會給你的日期不同，而且，時間，分鐘，小時，第二和等

date_default_timezone_set('America/Los_Angeles'); 
$now = '2016-10-13'; 
$returnDate = '2016-10-20'; 
$start = date_create($returnDate); 
$end = date_create($now); 
$diff=date_diff($end,$start); 
print_r($diff); 

DEMO

Answer 4

從手動

$pickupDate = new DateTime('2016-10-13'); 
$returnDate = new DateTime('2016-10-20'); 
$interval = $pickupDate->diff($returnDate); 
echo $interval->format('%R%a days'); 

http://php.net/manual/en/datetime.diff.php

Answer 5

date_default_timezone_set('America/Los_Angeles'); 

$pickupDate = '2016-10-13'; 
$returnDate = '2016-10-20'; //use signle quote same as pickupDate 

$diff = abs(strtotime($returnDate) - strtotime($pickupDate)); // change the order 
$years = floor($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)); 

printf("%d years, %d months, %d days\n", $years, $months, $days); 

感謝