如何使用PHP找到两个日期之间的天数?


当前回答

出于类似的目的,我在我的作曲项目中使用Carbon。

就像这样简单:

$dt = Carbon::parse('2010-01-01');
echo $dt->diffInDays(Carbon::now());

其他回答

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;

}

function howManyDays($startDate,$endDate) {

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

return $res;
} 
$early_start_date = date2sql($_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;
}

出于类似的目的,我在我的作曲项目中使用Carbon。

就像这样简单:

$dt = Carbon::parse('2010-01-01');
echo $dt->diffInDays(Carbon::now());
$now = time(); // or your date as well
$your_date = strtotime("2010-01-31");
$datediff = $now - $your_date;

echo round($datediff / (60 * 60 * 24));