Calculate working days between to dates

Question

I want to calculate the number of working days.

By using carbon I can calculate days and reduce weekends.

$num_days = $to_date->diffInWeekdays($from_date) + 1;

And I have an array of holidays, and I want to reduce the number of days if there is a holiday in between the days.

Is there any way to do this.

Thank you

Govern

Are the dates in your holiday array just strings or Carbon instances? If they're strings how are they formatted? – Stichous

Answer 1

You could use diffInDaysFiltered to achieve what you're after.

Assuming your holidays are an array of Carbon instances you could do something like:

$start = Carbon::now()->setDate(2014, 1, 1);
$end = Carbon::now()->setDate(2015, 1, 1);

$holidays = [
    Carbon::create(2014, 2, 2),
    Carbon::create(2014, 4, 17),
    Carbon::create(2014, 5, 19),
    Carbon::create(2014, 7, 3),
];

$days = $start->diffInDaysFiltered(function (Carbon $date) use ($holidays) {
    return $date->isWeekday() && !in_array($date, $holidays);
}, $end);

If it's just an array of strings then you could do something like:

!in_array($date->format('[the-format-of-your-dates]'), $holidays)

Stichous

This is a nice and short way to do it. I just would add "false" as the third parameter of diffInDaysFiltered to get negative values for a past date. Otherwise you will get a positive value when the $end date is in the past or future. $days = $start->diffInDaysFiltered(function (Carbon $date) use ($holidays) { return $date->isWeekday() && !in_array($date, $holidays); }, $end,false); – Lingle

This works fine, though it seems not to be skipping holidays. Please guide Thanks – Mendiola

@MasabaJamesMoses I'd recommend you create a new question as there are potentially far too many reasons why this might not be working for you. – Stichous

Answer 2

Here is a small utility function to calculate working days:

function getWorkingDaysCount($from, $to) {
    $workingDays = [1, 2, 3, 4, 5];         // Working days (week days)
    $holidayDays = ['*-12-25', '*-01-01'];  // Holidays array, add desired dates to this array 

    $from = new DateTime($from);
    $to = new DateTime($to);
    $to->modify('+1 day');
    $interval = new DateInterval('P1D');
    $periods = new DatePeriod($from, $interval, $to);

    $days = 0;
    foreach ($periods as $period) {
      if (!in_array($period->format('N'), $workingDays)) continue;
      if (in_array($period->format('Y-m-d'), $holidayDays)) continue;
      if (in_array($period->format('*-m-d'), $holidayDays)) continue;
      $days++;
  }
  return $days;
}

You can pass the starting and end date to get the number of days.

Homogeneous

Answer 3

You can take advantage of Carbon's diffInDaysFiltered() and isWeekend() methods. Below is a modified code from Carbon's documentation.

$daysForExtraCoding = $dt->diffInDaysFiltered(function(Carbon $date) {
   return !$date->isWeekend();
}, $dt2);

echo $daysForExtraCoding;      // 260

Shorter code

$daysForExtraCoding = $dt->diffInWeekdays($dt2);

Linette

Answer 4

You can get this by following code

<?php
    function workdays($start,$end,$holidays){
        $end = strtotime($end);
        $start = strtotime($start);
        $days = ($end - $start) / 86400 + 1;

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

        $the_first_day_of_week = date("N", $start);
        $the_last_day_of_week = date("N", $end);

        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) $no_remaining_days--;
        }
        else {
            if ($the_first_day_of_week == 7) {
                $no_remaining_days--;

                if ($the_last_day_of_week == 6) {
                    $no_remaining_days--;
                }
            }
            else {
                $no_remaining_days -= 2;
            }
        }

       $workingDays = $no_full_weeks * 5;
        if ($no_remaining_days > 0 )
        {
          $workingDays += $no_remaining_days;
        }

        foreach($holidays as $holiday){
            $time_stamp=strtotime($holiday);
            if ($start <= $time_stamp && $time_stamp <= $end && date("N",$time_stamp) != 6 && date("N",$time_stamp) != 7)
                $workingDays--;
        }

        return $workingDays;
    }



    echo workdays("2016-01-04","2016-01-30",array("2016-01-25","2016-01-26","2016-01-27"))

    ?>

Sandrocottus

Answer 5

function getWorkingDays($days1,$start_dt,$end_dt,$holidays) {


    $date1 = $start_dt;
    $date2 = $end_dt;

    $array = array();
    $query = $holidays;// pass array of holidays.
    foreach($query as $row)
    {
        $array[] = $row['h_date'];
    }

    $workingDays = [1, 2, 3, 4, 5];// 1 = Mon,...
    $holidayDays = $array; # variable and fixed holidays

    $from = new DateTime($date1);
    $to = new DateTime($date2);
    $to->modify('+1 day');
    $interval = new DateInterval('P1D');
    $periods = new DatePeriod($from, $interval, $to);

    $nameOfDay = date('N', strtotime($date1));
    if($nameOfDay==7)
    {
        $nameOfDay=0;
    }
    $days = $nameOfDay;
    $fullDate=explode('-',$date1);

    $y=$fullDate['0'];
    $m=$fullDate['1'];
    $d=$fullDate['2'];

    foreach ($periods as $period) {
        if (!in_array($period->format('N'), $workingDays)) {

            $d++;
            continue;
        }
        if (in_array($period->format('Y-m-d'), $holidayDays)) {
            if($d==32 && $m==12)
            {
                $d=1;
                $m=1;
                $y++;
            }

            $d++;
            continue;
        }
        if (in_array($period->format('*-m-d'), $holidayDays)) {

            $d++;
            continue;
        }

        switch ($m) {

            case '2':
                if( (0 == $y % 4) and (0 != $y % 100) or (0 == $y % 400) )
                {
                    if($d>=30)
                    {
                        $d=1;
                        $m++;
                        break;
                    }  
                }
                else
                {
                    if($d>=29)
                    {
                        $d=1;
                        $m++;
                        break;
                    }  
                }

            case '4':
            case '6':
            case '9':
            case '11':
                if($d>=31)
                {
                    $d=1;
                    $m++;
                }
                break;

            case '1':
            case '3':
            case '5':
            case '7':
            case '8':
            case '10':
                if($d>=32)
                {
                    $d=1;
                    $m++;   
                }
                break;

            case '12':
                if($d>=32)
                {
                    $d=1;
                    $m=1;
                    $y++;
                }
                break;
        }

        $dn = date('D', strtotime($d.'-'.$m.'-'.$y));
        if($d==1 && $dn=='Sun')
        {
            $d++;
        }
        if($d==1 && $dn=='Sat')
        {
            $d=$d+2;
        }

        $dayName = date('D', strtotime($d.'-'.$m.'-'.$y));

        $planbookid = $this->planbook->get_planbook_id();

        $d++;
        $days++;
    }
    echo json_encode($result);
}

Laurinda