递归函数不返回预期结果PHP [重复]

Question

我正在尝试创建一个递归函数，该函数根据一些预定义的规则计算礼物送达日期。 1. 预订一天后即可送出礼物 2. 如果在周六或周日预订，礼品将在下一个工作日 + 1 个处理日后送达。 3. 结果日期可能不在预定义的假期中。

我创建了以下函数，但它返回的日期不正确。

//The delivery date might not be from these dates
$holidays_selected = array('23-10-2015','24-10-2015','28-10-2015');

echo $gift_delivery_date = getGiftDeliveryDate(date('d-m-Y',strtotime('+1 Day')),$holidays_selected);
//It prints 25-10-2015 what i expect is 27-10-2015

function getGiftDeliveryDate($asuumed_date,$holidays){
        $tomorrow = '';
        if (in_array($asuumed_date,$holidays)) {
            $tomorrow = date('d-m-Y',strtotime($asuumed_date.'+1 Day'));
            getGiftDeliveryDate($tomorrow,$holidays);

        } else if (date('N',strtotime($asuumed_date)) == 6) {
            $tomorrow = date('d-m-Y',strtotime($asuumed_date.'+3 Day'));
                if(in_array($tomorrow,$holidays)){
                    $tomorrow = date('d-m-Y',strtotime($tomorrow.'+1 Day'));
                    getGiftDeliveryDate($tomorrow,$holidays);
                }
        } else if (date('N',strtotime($asuumed_date)) == 7) {
            $tomorrow = date('d-m-Y',strtotime($asuumed_date.'+2 Day'));
                if(in_array($tomorrow,$holidays)){
                    $tomorrow = date('d-m-Y',strtotime($tomorrow.'+1 Day'));
                    getGiftDeliveryDate($tomorrow,$holidays);
                }
        } else {
            $tomorrow = $asuumed_date;
        }

    return $tomorrow;
}

编辑

我期望的输出是27-10-2015，但它给出了25-10-2015 作为最终输出

Answer 1

你错过了函数的返回值

function getGiftDeliveryDate($asuumed_date, $holidays) {
    if (in_array($asuumed_date, $holidays)) {
        $tomorrow = date('d-m-Y', strtotime($asuumed_date . '+1 Day'));
        <b>$tomorrow =</b> getGiftDeliveryDate($tomorrow, $holidays);

    } else if (date('N', strtotime($asuumed_date)) == 6) {
        $tomorrow = date('d-m-Y', strtotime($asuumed_date . '+3 Day'));
        if (in_array($tomorrow, $holidays)) {
            $tomorrow = date('d-m-Y', strtotime($tomorrow . '+1 Day'));
            <b>$tomorrow =</b> getGiftDeliveryDate($tomorrow, $holidays);
        }
    } else if (date('N', strtotime($asuumed_date)) == 7) {
        $tomorrow = date('d-m-Y', strtotime($asuumed_date . '+2 Day'));
        if (in_array($tomorrow, $holidays)) {
            $tomorrow = date('d-m-Y', strtotime($tomorrow . '+1 Day'));
            <b>$tomorrow =</b> getGiftDeliveryDate($tomorrow, $holidays);
        }
    } else {
        $tomorrow = $asuumed_date;
    }

    return $tomorrow;
}