选择的答案不是最正确的答案，因为它将在UTC之外失败。 根据时区(列表)，可能会有时间调整创建“没有”24小时的日子，这将使计算(60*60*24)失败。

这里有一个例子:

date_default_timezone_set('europe/lisbon');
$time1 = strtotime('2016-03-27');
$time2 = strtotime('2016-03-29');
echo floor( ($time2-$time1) /(60*60*24));
 ^-- the output will be **1**

因此，正确的解决方案是使用DateTime

date_default_timezone_set('europe/lisbon');
$date1 = new DateTime("2016-03-27");
$date2 = new DateTime("2016-03-29");

echo $date2->diff($date1)->format("%a");
 ^-- the output will be **2**

function get_daydiff($end_date,$today)
{
    if($today=='')
    {
        $today=date('Y-m-d');
    }
    $str = floor(strtotime($end_date)/(60*60*24)) - floor(strtotime($today)/(60*60*24));
    return $str;
}
$d1 = "2018-12-31";
$d2 = "2018-06-06";
echo get_daydiff($d1, $d2);

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

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);

你可以试试下面的代码:

$dt1 = strtotime("2019-12-12"); //Enter your first date
$dt2 = strtotime("12-12-2020"); //Enter your second date
echo abs(($dt1 - $dt2) / (60 * 60 * 24));

function howManyDays($startDate,$endDate) {

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

return $res;
}