您还可以使用以下代码通过向上舍入分数来返回日期差异$date1=$duedate；//指定到期日echo$date2=日期（“Y-m-d”）；//当前日期$ts1=字符串时间（$date1）；$ts2=字符串时间（$date2）；$seconds_diff=$ts1-$ts2；echo$datediff=ceil（（$seconds_diff/3600）/24）；//天内返回

如果您使用php的floor方法而不是ceil，它将返回舍入分数。请检查此处的差异，有时，如果您的临时服务器时区与现场站点时区不同，在这种情况下，您可能会得到不同的结果，因此请相应地更改条件。

我在PHP5.2中遇到了同样的问题，并用MySQL解决了这个问题。可能并不是你想要的，但这会奏效，并返回天数：

$datediff_q = $dbh->prepare("SELECT DATEDIFF(:date2, :date1)");
$datediff_q->bindValue(':date1', '2007-03-24', PDO::PARAM_STR);
$datediff_q->bindValue(':date2', '2009-06-26', PDO::PARAM_STR);
$datediff = ($datediff_q->execute()) ? $datediff_q->fetchColumn(0) : false;

此处有更多信息http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff

当PHP 5.3（分别为date_diff（））不可用时，我使用了我编写的以下函数：

        function dateDifference($startDate, $endDate)
        {
            $startDate = strtotime($startDate);
            $endDate = strtotime($endDate);
            if ($startDate === false || $startDate < 0 || $endDate === false || $endDate < 0 || $startDate > $endDate)
                return false;

            $years = date('Y', $endDate) - date('Y', $startDate);

            $endMonth = date('m', $endDate);
            $startMonth = date('m', $startDate);

            // Calculate months
            $months = $endMonth - $startMonth;
            if ($months <= 0)  {
                $months += 12;
                $years--;
            }
            if ($years < 0)
                return false;

            // Calculate the days
            $measure = ($months == 1) ? 'month' : 'months';
            $days = $endDate - strtotime('+' . $months . ' ' . $measure, $startDate);
            $days = date('z', $days);   

            return array($years, $months, $days);
        }

对于php版本>=5.3：创建两个日期对象，然后使用date_diff（）函数。它将返回phpDateInterval对象。请参阅文档

$date1=date_create("2007-03-24");
$date2=date_create("2009-06-26");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");

我更喜欢使用date_create和date_diff对象。

代码：

$date1 = date_create("2007-03-24");
$date2 = date_create("2009-06-26");

$dateDifference = date_diff($date1, $date2)->format('%y years, %m months and %d days');

echo $dateDifference;

输出：

2 years, 3 months and 2 days

有关更多信息，请阅读PHP date_diff手册

根据手册date_diff是的别名日期时间：：diff（）

我有一些简单的逻辑：

<?php
    per_days_diff('2011-12-12','2011-12-29')
    function per_days_diff($start_date, $end_date) {
        $per_days = 0;
        $noOfWeek = 0;
        $noOfWeekEnd = 0;
        $highSeason=array("7", "8");

        $current_date = strtotime($start_date);
        $current_date += (24 * 3600);
        $end_date = strtotime($end_date);

        $seassion = (in_array(date('m', $current_date), $highSeason))?"2":"1";

        $noOfdays = array('');

        while ($current_date <= $end_date) {
            if ($current_date <= $end_date) {
                $date = date('N', $current_date);
                array_push($noOfdays,$date);
                $current_date = strtotime('+1 day', $current_date);
            }
        }

        $finalDays = array_shift($noOfdays);
        //print_r($noOfdays);
        $weekFirst = array("week"=>array(),"weekEnd"=>array());
        for($i = 0; $i < count($noOfdays); $i++)
        {
            if ($noOfdays[$i] == 1)
            {
                //echo "This is week";
                //echo "<br/>";
                if($noOfdays[$i+6]==7)
                {
                    $noOfWeek++;
                    $i=$i+6;
                }
                else
                {
                    $per_days++;
                }
                //array_push($weekFirst["week"],$day);
            }
            else if($noOfdays[$i]==5)
            {
                //echo "This is weekend";
                //echo "<br/>";
                if($noOfdays[$i+2] ==7)
                {
                    $noOfWeekEnd++;
                    $i = $i+2;
                }
                else
                {
                    $per_days++;
                }
                //echo "After weekend value:- ".$i;
                //echo "<br/>";
            }
            else
            {
                $per_days++;
            }
        }

        /*echo $noOfWeek;
          echo "<br/>";
          echo $noOfWeekEnd;
          echo "<br/>";
          print_r($per_days);
          echo "<br/>";
          print_r($weekFirst);
        */

        $duration = array("weeks"=>$noOfWeek, "weekends"=>$noOfWeekEnd, "perDay"=>$per_days, "seassion"=>$seassion);
        return $duration;
      ?>