Function to Get Number of Weekdays Between Two Dates Excluding Holidays

$(document).ready(() => {
  $('#calc').click(() => {
  var d1 = $('#d1').val();
  var d2 = $('#d2').val();
    $('#dif').text(workingDaysBetweenDates(d1,d2));
  });
});

let workingDaysBetweenDates = (d0, d1) => {
  /* Two working days and an sunday (not working day) */
  var holidays = ['2016-05-03', '2016-05-05', '2016-05-07'];
  var startDate = parseDate(d0);
  var endDate = parseDate(d1);  

// Validate input
  if (endDate <= startDate) {
    return 0;
  }

// Calculate days between dates
  var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
  startDate.setHours(0, 0, 0, 1);  // Start just after midnight
  endDate.setHours(23, 59, 59, 999);  // End just before midnight
  var diff = endDate - startDate;  // Milliseconds between datetime objects    
  var days = Math.ceil(diff / millisecondsPerDay);

  // Subtract two weekend days for every week in between
  var weeks = Math.floor(days / 7);
  days -= weeks * 2;

  // Handle special cases
  var startDay = startDate.getDay();
  var endDay = endDate.getDay();
    
  // Remove weekend not previously removed.   
  if (startDay - endDay > 1) {
    days -= 2;
  }
  // Remove start day if span starts on Sunday but ends before Saturday
  if (startDay == 0 && endDay != 6) {
    days--;  
  }
  // Remove end day if span ends on Saturday but starts after Sunday
  if (endDay == 6 && startDay != 0) {
    days--;
  }
  /* Here is the code */
  holidays.forEach(day => {
    if ((day >= d0) && (day <= d1)) {
      /* If it is not saturday (6) or sunday (0), substract it */
      if ((parseDate(day).getDay() % 6) != 0) {
        days--;
      }
    }
  });
  return days;
}
           
function parseDate(input) {
    // Transform date from text to date
  var parts = input.match(/(\d+)/g);
  // new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
  return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="d1" value="2016-05-02"><br>
<input type="text" id="d2" value="2016-05-08">

<p>Working days count: <span id="dif"></span></p>
<button id="calc">Calc</button>

<p>
Now it shows 5 days, but I need for example add holidays 
3 and 5 May (2016-05-03 and 2016-05-05) so the result will be 3 working days
</p>

How to find a number of workdays between two dates in Python?

use the numpy function busday_count:

from datetime import date
import numpy as np

d1 = date(2019, 3, 1)
d2 = date(2019, 5, 6)

days = np.busday_count( d1, d2 )
print (days)

or

from datetime import date,timedelta
d1 = date(2019, 3, 1)
d2 = date(2019, 5, 6)
daygenerator = (d1 + timedelta(x + 1) for x in range((d2 - d1).days)) # generate all days from d1 to d2
print (sum(1 for day in daygenerator if day.weekday() < 5)) 

Calculate the number of weekdays between 2 dates in R

Date1 <- as.Date("2011-01-30")
Date2 <- as.Date("2011-02-04")    
sum(!weekdays(seq(Date1, Date2, "days")) %in% c("Saturday", "Sunday"))

EDIT: And Zach said, let there be Vectorize :)

Dates1 <- as.Date("2011-01-30") + rep(0, 10)
Dates2 <- as.Date("2011-02-04") + seq(0, 9)
Nweekdays <- Vectorize(function(a, b) 
  sum(!weekdays(seq(a, b, "days")) %in% c("Saturday", "Sunday")))
Nweekdays(Dates1, Dates2)

Calculate business days

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
?>

Find day difference between two dates (excluding weekend days)

Maybe someone else can help you converting this function into JQuery's framework...

I found this function here.

function calcBusinessDays(dDate1, dDate2) { // input given as Date objects

  var iWeeks, iDateDiff, iAdjust = 0;

  if (dDate2 < dDate1) return -1; // error code if dates transposed

  var iWeekday1 = dDate1.getDay(); // day of week

  var iWeekday2 = dDate2.getDay();

  iWeekday1 = (iWeekday1 == 0) ? 7 : iWeekday1; // change Sunday from 0 to 7

  iWeekday2 = (iWeekday2 == 0) ? 7 : iWeekday2;

  if ((iWeekday1 > 5) && (iWeekday2 > 5)) iAdjust = 1; // adjustment if both days on weekend

  iWeekday1 = (iWeekday1 > 5) ? 5 : iWeekday1; // only count weekdays

  iWeekday2 = (iWeekday2 > 5) ? 5 : iWeekday2;

  // calculate differnece in weeks (1000mS * 60sec * 60min * 24hrs * 7 days = 604800000)

  iWeeks = Math.floor((dDate2.getTime() - dDate1.getTime()) / 604800000)

  if (iWeekday1 < iWeekday2) { //Equal to makes it reduce 5 days

    iDateDiff = (iWeeks * 5) + (iWeekday2 - iWeekday1)

  } else {

    iDateDiff = ((iWeeks + 1) * 5) - (iWeekday1 - iWeekday2)

  }

  iDateDiff -= iAdjust // take into account both days on weekend

  return (iDateDiff + 1); // add 1 because dates are inclusive

}

var date1 = new Date("August 11, 2010 11:13:00");

var date2 = new Date("August 16, 2010 11:13:00");

alert(calcBusinessDays(date1, date2));

C# How to calculate working days excluding specific dates (Holidays)

/// <summary> Get working days between two dates (Excluding a list of dates - Holidays) </summary>
/// <param name="dtmCurrent">Current date time</param>
/// <param name="dtmFinishDate">Finish date time</param>
/// <param name="lstExcludedDates">List of dates to exclude (Holidays)</param>
public static int fwGetWorkingDays(this DateTime dtmCurrent, DateTime dtmFinishDate, List<DateTime> lstExcludedDates)
{
    Func<DateTime, bool> workDay = currentDate =>
            (
                currentDate.DayOfWeek == DayOfWeek.Saturday ||
                currentDate.DayOfWeek == DayOfWeek.Sunday ||
                lstExcludedDates.Exists(evalDate => evalDate.Date.Equals(currentDate.Date))
            );

    return Enumerable.Range(0, 1 + (dtmFinishDate - dtmCurrent).Days).Count(intDay => workDay(dtmCurrent.AddDays(intDay)));
}

I created this extension that basically takes two dates and a list of dates to exclude (Holidays).
Surprisingly this method is very fast and has a validation to exclude the time from the DateTime when compare against the list of exclusions.

MySQL function to find the number of working days between two dates

This expression -

5 * (DATEDIFF(@E, @S) DIV 7) + MID('0123444401233334012222340111123400012345001234550', 7 * WEEKDAY(@S) + WEEKDAY(@E) + 1, 1)

calculates the number of business days between the start date @S and the end date @E.

Assumes end date (@E) is not before start date (@S).
Compatible with DATEDIFF in that the same start date and end date
gives zero business days.
Ignores holidays.

The string of digits is constructed as follows. Create a table of
start days and end days, the rows must start with monday (WEEKDAY
0) and the columns must start with Monday as well. Fill in the
diagonal from top left to bottom right with all 0 (i.e. there are 0
working days between Monday and Monday, Tuesday and Tuesday, etc.).
For each day start at the diagonal (must always be 0) and fill in
the columns to the right, one day at a time. If you land on a
weekend day (non business day) column, the number of business days
doesn't change, it is carried from the left. Otherwise, the number
of business days increases by one. When you reach the end of the
row loop back to the start of the same row and continue until you
reach the diagonal again. Then go on to the next row.

E.g. Assuming Saturday and Sunday are not business days -

 | M T W T F S S
-|--------------
M| 0 1 2 3 4 4 4
T| 4 0 1 2 3 3 3
W| 3 4 0 1 2 2 2
T| 2 3 4 0 1 1 1
F| 1 2 3 4 0 0 0
S| 1 2 3 4 5 0 0
S| 1 2 3 4 5 5 0

Then concatenate the 49 values in the table into the string.

Please let me know if you find any bugs.

-Edit
improved table:

 | M T W T F S S
-|--------------
M| 0 1 2 3 4 4 4
T| 4 0 1 2 3 3 3
W| 3 4 0 1 2 2 2
T| 2 3 4 0 1 1 1
F| 1 2 3 4 0 0 0
S| 0 1 2 3 4 0 0
S| 0 1 2 3 4 4 0

improved string: '0123444401233334012222340111123400001234000123440'

improved expression:

5 * (DATEDIFF(@E, @S) DIV 7) + MID('0123444401233334012222340111123400001234000123440', 7 * WEEKDAY(@S) + WEEKDAY(@E) + 1, 1)

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