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;

}

最简单的方法来找出两个日期之间的天数差

$date1 = strtotime("2019-05-25"); 
$date2 = strtotime("2010-06-23");

$date_difference = $date2 - $date1;

$result =  round( $date_difference / (60 * 60 * 24) );

echo $result;

使用这个:)

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

现在起作用了

$early_start_date = date2sql($_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;
}

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!