How to get the total working days of the given month without function

Question

How can i get the total number of working days in the month without using function

function countDays($year, $month, $ignore) {
    $count = 0;
    $counter = mktime(0, 0, 0, $month, 1, $year);
    while (date("n", $counter) == $month) {
        if (in_array(date("w", $counter), $ignore) == false) {
            $count++;
        }
        $counter = strtotime("+1 day", $counter);
    }
    return $count;
}
echo countDays(2013, 1, array(0, 6)); // 23

In the above code found function is used..

But how can i use it without using function ?

Answer 1

Here's a function from the user comments on the date() function page in the PHP manual. It's an improvement of an earlier function in the comments that adds support for leap years.

Enter the starting and ending dates, along with an array of any holidays that might be in between, and it returns the working days as an integer:

---

<?php
    //The function returns the no. of business days between two dates and it skips the holidays
    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;
    }

    //Example:

    $holidays=array("2008-12-25","2008-12-26","2009-01-01");

    echo getWorkingDays("2008-12-22","2009-01-02",$holidays)
    // => will return 7
    ?>

Answer 2

$numdays = cal_days_in_month (CAL_GREGORIAN, $mon,$yr);
for($i=1;$i<=$numdays;$i++)
    {
    if(date('N',strtotime($y.'-'.$m.'-'.$i))==7)
    $sun++;
    }

Sunday calculation after u calculate employee working days

$tot=$numdays-$sun-$emp_working_days;echo $tot;

Answer 3

I don't know what you want exactly. But, Could it be this?

$year = 2015;
$month = 3;
$count = 0;
$counter = mktime(0, 0, 0, $month, 1, $year);
while (date("n", $counter) == $month) {
    if (in_array(date("w", $counter), $ignore) == false) {
        $count++;
    }
    $counter = strtotime("+1 day", $counter);
}

echo $count;