Question

我有一个最奇怪的问题。我有一个脚本，用于计算当月的当前工作日以及当月的工作日总量。例如，今天6月25日，脚本将输出“第17/21天”。

奇怪的是，从第12天到月底，在第12个星期一的月份，当前的工作日显示比实际工作日少0.0417天。例如，3月11日（周日）它显示“7/22”。但是在3月12日，它显示“7.95833333 / 22”

下面是我在互联网上某个帖子上找到的脚本代码。除了这个奇怪的场合，它每隔一段时间都能完美运作。

//##########################################################
//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);

Answer 1

如果你正在做/ 86400件事，那么你将遇到闰秒问题。 http://en.wikipedia.org/wiki/Leap_second

最好使用类似strtotime('+1 day', $timestamp)或DateTime类的内容来日常移动或比较日期。

至少，如果你要处理一整天的增量，只需将结果四舍五入。

用PHP计算工作天数 - 在星期一12日的月份不起作用

1 个答案: