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

    // Change this to the day in the future
$day = 15;

// Change this to the month in the future
$month = 11;

// Change this to the year in the future
$year = 2012;

// $days is the number of days between now and the date in the future
$days = (int)((mktime (0,0,0,$month,$day,$year) - time(void))/86400);

echo "There are $days days until $day/$month/$year";

function howManyDays($startDate,$endDate) {

    $date1  = strtotime($startDate." 0:00:00");
    $date2  = strtotime($endDate." 23:59:59");
    $res    =  (int)(($date2-$date1)/86400);        

return $res;
} 

我已经尝试了答案中几乎所有的方法。但是DateTime和date_create在所有测试用例中都没有给出正确答案。特别在2月和3月或12月和1月进行测试。

所以，我想出了混合溶液。

public static function getMonthsDaysDiff($fromDate, $toDate, $includingEnding = false){
    $d1=new DateTime($fromDate);
    $d2=new DateTime($toDate);
    if($includingEnding === true){
        $d2 = $d2->modify('+1 day');
    }
    $diff = $d2->diff($d1);
    $months = (($diff->format('%y') * 12) + $diff->format('%m'));

    $lastSameDate = $d1->modify("+$months month");
    $days = date_diff(
        date_create($d2->format('Y-m-d')),
        date_create($lastSameDate->format('Y-m-d'))
    )->format('%a');

    $return = ['months' => $months,
        'days' => $days];
}

我知道，性能方面这是相当昂贵的。你也可以把它扩展到年限。

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!

$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;

echo round($datediff / (60 * 60 * 24));