我阅读了所有以前的解决方案，没有一个使用PHP 5.3工具:DateTime::Diff和DateInterval::Days

DateInterval::Days精确地包含日期之间的天数。没有必要创造一些特别和奇异的东西。

/**
 * We suppose that PHP is configured in UTC
 * php.ini configuration:
 * [Date]
 * ; Defines the default timezone used by the date functions
 * ; http://php.net/date.timezone
 * date.timezone = UTC
 * @link http://php.net/date.timezone
 */

/**
 * getDaysBetween2Dates
 *
 * Return the difference of days between $date1 and $date2 ($date1 - $date2)
 * if $absolute parameter is false, the return value is negative if $date2 is after than $date1
 *
 * @param DateTime $date1
 * @param DateTime $date2
 * @param Boolean $absolute
 *            = true
 * @return integer
 */
function getDaysBetween2Dates(DateTime $date1, DateTime $date2, $absolute = true)
{
    $interval = $date2->diff($date1);
    // if we have to take in account the relative position (!$absolute) and the relative position is negative,
    // we return negatif value otherwise, we return the absolute value
    return (!$absolute and $interval->invert) ? - $interval->days : $interval->days;
}

echo '<h3>2020-03-01 - 2020-02-01: 29 days as it\'s a standard leap year</h3>';
echo getDaysBetween2Dates(new DateTime("2020-03-01"), new DateTime("2020-02-01"), false);

echo '<h3>1900-03-01 - 1900-02-01: 28 days as it\'s a "standard" century</h3>';
echo getDaysBetween2Dates(new DateTime("1900-03-01"), new DateTime("1900-02-01"), false);

echo '<h3>2000-03-01 - 2000-02-01: 29 days as it\'s a century multiple of 400: 2000=400x5</h3>';
echo getDaysBetween2Dates(new DateTime("2000-03-01"), new DateTime("2000-02-01"), false);

echo '<h3>2020-03-01 - 2020-04-01: -28 days as 2020-03-01 is before 2020-04-01</h3>';
echo getDaysBetween2Dates(new DateTime("2020-02-01"), new DateTime("2020-03-01"), false);

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

计算两个日期的差值:

$date1=date_create("2013-03-15");
$date2=date_create("2013-12-12");

$diff=date_diff($date1,$date2);

echo $diff->format("%R%a days");

输出: + 272天

函数的作用是:返回两个DateTime对象之间的差值。

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)

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。

使用这个:)

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

现在起作用了