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)

如果你正在使用MySql

function daysSince($date, $date2){
$q = "SELECT DATEDIFF('$date','$date2') AS days;";
$result = execQ($q);
$row = mysql_fetch_array($result,MYSQL_BOTH);
return ($row[0]);

}

function execQ($q){
$result = mysql_query( $q);
if(!$result){echo ('Database error execQ' . mysql_error());echo $q;}    
return $result;

}

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);

$early_start_date = date2sql($_POST['early_leave_date']);


$date = new DateTime($early_start_date);
$date->modify('+1 day');


$date_a = new DateTime($early_start_date . ' ' . $_POST['start_hr'] . ':' . $_POST['start_mm']);
$date_b = new DateTime($date->format('Y-m-d') . ' ' . $_POST['end_hr'] . ':' . $_POST['end_mm']);

$interval = date_diff($date_a, $date_b);


$time = $interval->format('%h:%i');
$parsed = date_parse($time);
$seconds = $parsed['hour'] * 3600 + $parsed['minute'] * 60;
//        display_error($seconds);

$second3 = $employee_information['shift'] * 60 * 60;

if ($second3 < $seconds)
    display_error(_('Leave time can not be greater than shift time.Please try again........'));
    set_focus('start_hr');
    set_focus('end_hr');
    return 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));