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私は最も奇妙な問題を抱えています。1 か月の現在の稼働日とその月の合計稼働日を計算するスクリプトがあります。たとえば、今日 6 月 25 日の場合、スクリプトは「day 17/21」と出力します。

奇妙なことに、12 日が月曜日である月では、12 日から月末まで、現在の稼働日は実際よりも 0.0417 日少なく表示されます。たとえば、3 月 11 日 (日曜日) の場合、「7/22」と表示されます。しかし、3 月 12 日の場合、「7.95833333/22」と表示されます。

以下は、インターネット上のどこかの投稿で見つけたスクリプトのコードです。この奇妙な機会を除いて、毎回完全に機能します。

//##########################################################
//Date Calculations
//##########################################################

    $datepicker = $_POST['datepicker'];
    $dpyear = date("Y", strtotime($datepicker));
    $dpmonth = date("m", strtotime($datepicker));
    $dpday = date("d", strtotime($datepicker));

        $end = date('Y/m/d', mktime(0, 0, 0, $dpmonth, $dpday, $dpyear));       // seconds * minutes * hours * days
        $first = date('Y/m/d', mktime(0, 0, 0, $dpmonth, 1, $dpyear));
        $firstOfYear = date('Y/m/d', mktime(0, 0, 0, 1, 1, $dpyear));
        $last = date('Y/m/d', mktime(0, 0, 0, $dpmonth + 1, 0, $dpyear));
        $firstprevyear = date('Y/m/d', mktime(0, 0, 0, $dpmonth, 1, $dpyear-1));
        $lastprevyear = date('Y/m/d', mktime(0, 0, 0, $dpmonth + 1, 0, $dpyear-1));
        $prevyearmonth = date('M\'y', mktime(0, 0, 0, $dpmonth + 1, 0, $dpyear-1));

        $prevyeardate = "sojd.InvoiceDate >= '$firstprevyear' and sojd.InvoiceDate <= '$lastprevyear' ";
        $mtddaterange = "sojd.InvoiceDate >= '$first' and sojd.InvoiceDate <= '$end' ";
        $ytddaterange = "sojd.InvoiceDate >= '$firstOfYear' and sojd.InvoiceDate <= '$end' ";



        if (date("w", strtotime($datepicker)) == 1) {       // if today is Monday, combine weekend and Monday's numbers
            $startDate = date('Y/m/d', mktime(0, 0, 0, $dpmonth, $dpday - 2, $dpyear));
            $prevdaterange = "sojd.InvoiceDate >= '$startDate' and sojd.InvoiceDate <= '$end'";
            $daterange = "sojd.InvoiceDate >= '$startDate' and sojd.InvoiceDate <= '$end'";
            $params = array($startDate, $end);
        } else {
            $prevdaterange = "sojd.InvoiceDate = '$end' ";
            $daterange = "sojd.InvoiceDate = '$end'";
            $params = array($end);
        }



    $holidays=array("2012-01-02","2012-05-28","2012-07-04","2012-09-03","2012-11-22","2012-12-25");

function getWorkingDays($startDate,$endDate,$holidays){
    // do strtotime calculations just once
    $endDate = strtotime($endDate);
    $startDate = strtotime($startDate);


    //The total number of days between the two dates. We compute the no. of seconds and divide it to 60*60*24
    //We add one to inlude both dates in the interval.
    $days = ($endDate - $startDate) / 86400 + 1;

    $no_full_weeks = floor($days / 7);
    $no_remaining_days = fmod($days, 7);

    //It will return 1 if it's Monday,.. ,7 for Sunday
    $the_first_day_of_week = date("N", $startDate);
    $the_last_day_of_week = date("N", $endDate);

    //---->The two can be equal in leap years when february has 29 days, the equal sign is added here
    //In the first case the whole interval is within a week, in the second case the interval falls in two weeks.
    if ($the_first_day_of_week <= $the_last_day_of_week) {
        if ($the_first_day_of_week <= 6 && 6 <= $the_last_day_of_week) $no_remaining_days--;
        if ($the_first_day_of_week <= 7 && 7 <= $the_last_day_of_week) $no_remaining_days--;
    }
    else {
        // (edit by Tokes to fix an edge case where the start day was a Sunday
        // and the end day was NOT a Saturday)

        // the day of the week for start is later than the day of the week for end
        if ($the_first_day_of_week == 7) {
            // if the start date is a Sunday, then we definitely subtract 1 day
            $no_remaining_days--;

            if ($the_last_day_of_week == 6) {
                // if the end date is a Saturday, then we subtract another day
                $no_remaining_days--;
            }
        }
        else {
            // the start date was a Saturday (or earlier), and the end date was (Mon..Fri)
            // so we skip an entire weekend and subtract 2 days
            $no_remaining_days -= 2;
        }
    }

    //The no. of business days is: (number of weeks between the two dates) * (5 working days) + the remainder
//---->february in none leap years gave a remainder of 0 but still calculated weekends between first and last day, this is one way to fix it
   $workingDays = $no_full_weeks * 5;
    if ($no_remaining_days > 0 )
    {
      $workingDays += $no_remaining_days;
    }

    //We subtract the holidays
    foreach($holidays as $holiday){
        $time_stamp=strtotime($holiday);
        //If the holiday doesn't fall in weekend
        if ($startDate <= $time_stamp && $time_stamp <= $endDate && date("N",$time_stamp) != 6 && date("N",$time_stamp) != 7)
            $workingDays--;
    }

    return $workingDays;
}
    $currSalesDay = getWorkingDays($first,$end,$holidays);
    $totalSalesDay = round(getWorkingDays($first,$last,$holidays),0);
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1 に答える 1

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何かをしている場合/ 86400、うるう秒の問題が発生します。http://en.wikipedia.org/wiki/Leap_second

strtotime('+1 day', $timestamp)日から日への移動や日付の比較には、または DateTime クラスのようなものを使用することをお勧めします。

少なくとも、1 日単位の増分を扱う場合は、結果を四捨五入してください。

于 2012-06-25T22:12:14.590 に答える