如何通过一次减去一天来找到最近的非假日、非周末日期？

Question

我有这样的对象数组：

$holiday_calendar = Array
(
    [0] => stdClass Object
        (
            [holiday] => New Year\'s Day
            [holidayDate] => 2018-01-01
        )

    [1] => stdClass Object
        (
            [holiday] => Republic Day
            [holidayDate] => 2018-01-26
        )

    [2] => stdClass Object
        (
            [holiday] => Holi
            [holidayDate] => 2018-03-02
        )
)

$payroll_days_list = Array
(
    [0] => stdClass Object
        (
            [payrollDay] => Mon
            [isPayrollDay] => Y
        )

    [1] => stdClass Object
        (
            [payrollDay] => Tue
            [isPayrollDay] => Y
        )

    [2] => stdClass Object
        (
            [payrollDay] => Wed
            [isPayrollDay] => Y
        )

    [3] => stdClass Object
        (
            [payrollDay] => Thu
            [isPayrollDay] => Y
        )

    [4] => stdClass Object
        (
            [payrollDay] => Fri
            [isPayrollDay] => Y
        )

    [5] => stdClass Object
        (
            [payrollDay] => Sat
            [isPayrollDay] => N
        )

    [6] => stdClass Object
        (
            [payrollDay] => Sun
            [isPayrollDay] => N
        )

)

我需要检查给定的日期是否不是假期，也不是周六或周日。

如果日期适逢节假日或星期六或星期日，则日期应该一次后移一天，直到它落在符合条件的日期。

我写过这段代码：

$dt = "04/03/2018";

echo $dt = $this->checkHolidayExists($dt, $holiday_calendar, $payroll_days_list);

function checkHolidayExists($dt, $holiday_calendar, $payroll_days_list) {
    if (empty($holiday_calendar) && empty($payroll_days_list)) {
        return $dt;
    } else {
        foreach ($holiday_calendar as $hc) {
            if ($hc->holidayDate == $dt) {
                $dt = date('Y-m-d', strtotime($dt . ' -1 days'));
                $this->checkHolidayExists($dt, $holiday_calendar, $payroll_days_list);
            }
        }

        foreach ($payroll_days_list as $pdl) {
            if ($pdl->payrollDay == date('D', strtotime($dt)) && $pdl->isPayrollDay == 'N') {
                $dt = date('Y-m-d', strtotime($dt . ' -1 days'));
                $this->checkHolidayExists($dt, $holiday_calendar, $payroll_days_list);
            }
        }
        return $dt;
    }
}

但是这个 returns 日期是 03/03/2018。

应该是01/03/2018因为4号是周日，3号是周六，2号是节假日

请让我知道我遗漏了什么。

Answer 1

我不得不填写一些缺失的组件，但也许您可以从我的 re-write 中收集必要的逻辑。我的 class 努力通过颠倒取消资格的对象数组的顺序来最大程度地减少递归。

代码：(Demo)

class PayDay
{
    public function __construct($Ymd = null) {
        $this->date = $Ymd ?: date("Y-m-d");

        $this->holiday_calendar = [
            (object)["holiday" => "New Year's Day", "holidayDate" => "2018-01-01"],
            (object)["holiday" => "Republic Day", "holidayDate" => "2018-01-26"],
            (object)["holiday" => "Holi", "holidayDate" => "2018-03-02"]
        ];

        $this->payroll_days_list = [
            (object)["payrollDay" => "Mon", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Tue", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Wed", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Thu", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Fri", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Sat", "isPayrollDay" => "N"],
            (object)["payrollDay" => "Sun", "isPayrollDay" => "N"]
        ];

        echo "Starting Date: ",$this->date,"\n";
    }

    public function getPayDate() {
        $hc_match = false;
        foreach (array_reverse($this->holiday_calendar) as $hc) {  // work in reverse to reduce recursions
            if ($hc->holidayDate == $this->date) {
                $hc_match = true;
                //echo "Avoided: ", $hc->holiday,"\n";
                $this->loseOneDay();
                //echo "New Date: ", $this->date,"\n";
            } elseif ($hc_match) {                            // break when consecutive adjustments cease, to reduce recursions
                break;
            }
        }

        $pdl_match = false;
        foreach (array_reverse($this->payroll_days_list) as $pdl) {  // work in reverse to reduce recursions
            if ($pdl->payrollDay == date("D", strtotime($this->date)) && $pdl->isPayrollDay == "N") {
                $pdl_match = true;
                //echo "Avoided: ", $pdl->payrollDay,"\n";
                $this->loseOneDay();
                //echo "New Date: ", $this->date,"\n";
            } elseif ($pdl_match) {                           // break when consecutive adjustments cease, to reduce recursions
                break;
            }
        }

        if ($hc_match || $pdl_match) {
            //echo "\tRECURSE\n";
            $this->getPayDate();
        }
        return $this->date;
    }

    public function loseOneDay(){
        return $this->date = date("Y-m-d", strtotime($this->date." -1 day"));
    }
}

$date = new PayDay("2018-03-04");
echo "Adjusted Date: ", $date->getPayDate();

输出：

Starting Date: 2018-03-04
Adjusted Date: 2018-03-01

请务必使您的日期格式自始至终保持一致，这样您就不会将 04/03/2018 与 2018-03-04 进行比较。当然，格式由你决定，为简单起见，$dt和$this->holiday_calendar->holidayDate选择一个即可。

至于您发布的代码，因为您的 foreach() 循环不包含 break，您在找到匹配后进行 extra/unnecessary 迭代。

我建议反向迭代对象数组（或者更好的是反向声明它们），以便您可以在同一个函数调用中执行连续调整。

再三考虑后，我想我可能会使用 DateTime 对象并花时间准备不合格的数据以简化迭代日期调整过程。下面的class根本就没有使用递归，而是一个dual-condition循环。（同样，如何导入取消资格的对象数组取决于您。）

代码：(Demo)

class PayDay
{
    public function __construct($dmY = null) {
        $this->date = (is_null($dmY) ? new DateTime("midnight") : DateTime::createFromFormat('!d/m/Y', $dmY));  // ! for midnight

        $holiday_calendar = [                                          // import this data however you wish
            (object)["holiday" => "New Year's Day", "holidayDate" => "2018-01-01"],
            (object)["holiday" => "Republic Day", "holidayDate" => "2018-01-26"],
            (object)["holiday" => "Holi", "holidayDate" => "2018-03-02"]
        ];
        $this->blacklist_holidates = [];
        if (!empty($holiday_calendar)) {
            foreach ($holiday_calendar as $obj) {
                $this->blacklist_holidates[$obj->holiday] = DateTime::createFromFormat('!Y-m-d', $obj->holidayDate);  // ! for midnight
            }
        }

        $payroll_days_list = [                                          // import this data however you wish
            (object)["payrollDay" => "Mon", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Tue", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Wed", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Thu", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Fri", "isPayrollDay" => "Y"],
            (object)["payrollDay" => "Sat", "isPayrollDay" => "N"],
            (object)["payrollDay" => "Sun", "isPayrollDay" => "N"]
        ];
        $this->blacklist_daynames = [];
        if (!empty($payroll_days_list)) {
            foreach ($payroll_days_list as $obj) {
                if ($obj->isPayrollDay == "N") {
                    $this->blacklist_daynames[] = $obj->payrollDay;
                }
            }
        }
    }

    public function getPayDate() {
        while (in_array($this->date, $this->blacklist_holidates) || in_array($this->date->format("D"), $this->blacklist_daynames)) {
            $this->date->sub(new DateInterval('P1D'));
        }
        return $this->date->format("d/m/Y");
    }
}

$input = "04/03/2018";
echo "Starting Date: $input\n";
$date = new PayDay($input);
echo "Qualifying PayDate: ",$date->getPayDate();

如何通过一次减去一天来找到最近的非假日、非周末日期？

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