这段代码为我工作，并用PHP 8版本测试:

function numberOfDays($startDate, $endDate) 
{
    //1) converting dates to timestamps
     $startSeconds = strtotime($startDate);
     $endSeconds = strtotime($endDate);
   
    //2) Calculating the difference in timestamps
    $diffSeconds = $startSeconds  - $endSeconds;
     
    
    //3) converting timestamps to days
    $days=round($diffSeconds / 86400);
    
      /*  note :
          1 day = 24 hours 
          24 * 60 * 60 = 86400 seconds
      */
   
    //4) printing the number of days
    printf("Difference between two dates: ". abs($days) . " Days ");
    
    return abs($days);
}

如果你使用的是PHP 5.3 >，这是目前为止最准确的计算绝对差值的方法:

$earlier = new DateTime("2010-07-06");
$later = new DateTime("2010-07-09");

$abs_diff = $later->diff($earlier)->format("%a"); //3

如果你需要一个相对的(带符号的)天数，可以用这个代替:

$earlier = new DateTime("2010-07-06");
$later = new DateTime("2010-07-09");

$pos_diff = $earlier->diff($later)->format("%r%a"); //3
$neg_diff = $later->diff($earlier)->format("%r%a"); //-3

更多关于php的DateInterval格式可以在这里找到:https://www.php.net/manual/en/dateinterval.format.php

$datediff = floor(strtotime($date1)/(60*60*24)) - floor(strtotime($date2)/(60*60*24));

如果需要的话:

$datediff=abs($datediff);

$early_start_date = date2sql($_POST['early_leave_date']);


$date = new DateTime($early_start_date);
$date->modify('+1 day');


$date_a = new DateTime($early_start_date . ' ' . $_POST['start_hr'] . ':' . $_POST['start_mm']);
$date_b = new DateTime($date->format('Y-m-d') . ' ' . $_POST['end_hr'] . ':' . $_POST['end_mm']);

$interval = date_diff($date_a, $date_b);


$time = $interval->format('%h:%i');
$parsed = date_parse($time);
$seconds = $parsed['hour'] * 3600 + $parsed['minute'] * 60;
//        display_error($seconds);

$second3 = $employee_information['shift'] * 60 * 60;

if ($second3 < $seconds)
    display_error(_('Leave time can not be greater than shift time.Please try again........'));
    set_focus('start_hr');
    set_focus('end_hr');
    return FALSE;
}

$start = '2013-09-08';
$end = '2013-09-15';
$diff = (strtotime($end)- strtotime($start))/24/3600; 
echo $diff;

使用这个:)

$days = (strtotime($endDate) - strtotime($startDate)) / (60 * 60 * 24);
print $days;

现在起作用了