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。

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);

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

$early_start_date = date2sql($_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;
}

如果你想在开始日期和结束日期之间重复所有的日子，我想出了这个:

$startdatum = $_POST['start']; // starting date
$einddatum = $_POST['eind']; // end date

$now = strtotime($startdatum);
$your_date = strtotime($einddatum);
$datediff = $your_date - $now;
$number = floor($datediff/(60*60*24));

for($i=0;$i <= $number; $i++)
{
    echo date('d-m-Y' ,strtotime("+".$i." day"))."<br>";
}