Question

我有代码：

$dateporngprakfin = "2015-02-01"

for($i=1;$i<=6;$i++){

            $dateporngprakfin = strtotime("+1 months", $dateporngprakfin);
            $datesum = date('Y-m-d', $dateporngprakfin);
            echo $datesum."<br>";
}

此代码将显示“2015-03-01”至“2015-08-01”，但我想按月计算日期。

Answer 1

使用cal_days_in_month（）函数。

$dateporngprakfin = date_create("2015-02-01");
    for($i=1;$i<=6;$i++){
        date_add($dateporngprakfin, date_interval_create_from_date_string('1 Months'));
        $countDays = cal_days_in_month(CAL_GREGORIAN, date_format($dateporngprakfin, 'm'), date_format($dateporngprakfin, 'Y'));        
        echo "<br />Count Days in ".date_format($dateporngprakfin, 'Y-m')." is ".$countDays;
    }

Answer 2

嗨试试这个以获得所需的日期

        $dateporngprakfin = "2013-12-31";
        $monthsToAdvance=1;
        function getDates($startDate, $monthsToAdvance)
        {
                    $dt = new DateTime($startDate);
                    $day = $dt->format('d');
                    $dt->setDate($dt->format('Y'),$dt->format('n'),1);
                    $dt->add(new DateInterval('P'.$monthsToAdvance.'M'));
                    $daysInMonth = cal_days_in_month(CAL_GREGORIAN,$dt->format('n'), $dt->format('Y'));
                    if($day > $daysInMonth) $day = $daysInMonth;
                    $dt->setDate($dt->format('Y'),$dt->format('n'),$day);
                    return $dt->format('Y-m-d');
        }
        echo getDates($startDate, $monthsToAdvance)

有关详细信息here

，请参阅此页面

Answer 3

您可以尝试此操作来计算一个月内的天数

$number = cal_days_in_month(CAL_GREGORIAN, 8, 2003);

它将返回2003年8月份的天数

每个月循环计算日期

3 个答案: