Question

我有一个记录集，它从SQL数据库中提取各种信息，包括日期。我想做的是计算从特定日期到今天的工作日数。我基本上知道自第一次约会以来已经过了多少个工作日。

我在工作日没有问题，这里有很多非常有用的帖子。我遇到的问题是我需要在数据库中填充的表中的每一行上重复。但由于每行的变量不同，在第一个结果之后页面是空白的，因为我假设脚本试图使用相同的命名变量再次运行。

有没有办法为这个总和包含php代码，以便它可以在下一行中使用相同的变量运行而不会影响前一行？

提前感谢您的帮助。

＆＃13;

<?php
//The function returns the no. of business days between two dates and it skips the holidays
function getWorkingDays($startDate,$endDate){
    // 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;
}
$approval = date('Y-m-d',strtotime($row_rsworksorders1['intraartapproval']));
$w_days = getWorkingDays($approval,(date('Y-m-d'))) . ' Days';

echo $w_days;
?>

＆＃13;

从数据库中提取的日期是$ row_rsworksorders1 [＆＃39; intraartapproval＆＃39;]）

显示记录集中两个日期的工作日结果PHP

0 个答案: