如果你想在开始日期和结束日期之间重复所有的日子，我想出了这个:

$startdatum = $_POST['start']; // starting date
$einddatum = $_POST['eind']; // end date

$now = strtotime($startdatum);
$your_date = strtotime($einddatum);
$datediff = $your_date - $now;
$number = floor($datediff/(60*60*24));

for($i=0;$i <= $number; $i++)
{
    echo date('d-m-Y' ,strtotime("+".$i." day"))."<br>";
}

如果你有以秒为单位的时间(即unix时间戳)，那么你可以简单地减去时间并除以86400(秒/天)

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

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);
}

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

如果需要的话:

$datediff=abs($datediff);

从PHP 5.3及以上版本开始，添加了新的日期/时间函数来获得不同:

$datetime1 = new DateTime("2010-06-20");

$datetime2 = new DateTime("2011-06-22");

$difference = $datetime1->diff($datetime2);

echo 'Difference: '.$difference->y.' years, ' 
                   .$difference->m.' months, ' 
                   .$difference->d.' days';

print_r($difference);

结果如下:

Difference: 1 years, 0 months, 2 days

DateInterval Object
(
    [y] => 1
    [m] => 0
    [d] => 2
    [h] => 0
    [i] => 0
    [s] => 0
    [invert] => 0
    [days] => 367
)

希望能有所帮助!