Question

我有两次，就像这样：

$date1 = strtotime("02/12/2019 10:10:54 am");  
     $date2 = strtotime("02/12/2019 10:11:07 pm");
     $diff = abs($date2 - $date1); 
     $years = floor($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)); 
    $hours = floor(($diff - $years * 365*60*60*24  
   - $months*30*60*60*24 - $days*60*60*24) 
                               / (60*60));
    $minutes = floor(($diff - $years * 365*60*60*24  
     - $months*30*60*60*24 - $days*60*60*24  
                  - $hours*60*60)/ 60);
  $seconds = floor(($diff - $years * 365*60*60*24  
 - $months*30*60*60*24 - $days*60*60*24 
        - $hours*60*60 - $minutes*60));
  printf(" %d days, %d hours, "
 . "%d minutes, %d seconds", 
         $days, $hours, $minutes, $seconds);

我想计算它们之间的差异，但这不起作用。

Answer 1

使用DateTime类要容易得多，该类具有一个diff方法，该方法可以产生一个DateInterval对象，该对象可以以您喜欢的任何格式输出差异：

$date1 = new DateTime("02/12/2019 10:10:54 am"); 
$date2 = new DateTime("02/12/2019 10:11:07 pm");
$diff = $date1->diff($date2);
echo $diff->format('%y years, %m months, %d days, %h hours, %i minutes and %s seconds');

输出：

0 years, 0 months, 0 days, 12 hours, 0 minutes and 13 seconds

Demo on 3v4l.org

时间为am和pm的两个时间之间的时差

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