查找两个日期之间的天差（不包括周末）

Question

Hi i am using jquery-ui datepicker to select date and date.js to find difference between 2 dates.

Right now the problem is I want to exclude weekend days from calculation (saturday and sunday). How should i do that?

For example the user select start date (13/8/2010) and end date (16/8/2010). Since 14/8/2010 and 15/8/2010 is in week days, instead of 4 days total, i want it to be only 2 days.

This is the code im using right now:

<script type="text/javascript">

    $("#startdate, #enddate").change(function() {       

    var d1 = $("#startdate").val();
    var d2 = $("#enddate").val();

            var minutes = 1000*60;
            var hours = minutes*60;
            var day = hours*24;

            var startdate1 = getDateFromFormat(d1, "d-m-y");
            var enddate1 = getDateFromFormat(d2, "d-m-y");

            var days = 1 + Math.round((enddate1 - startdate1)/day);             

    if(days>0)
    { $("#noofdays").val(days);}
    else
    { $("#noofdays").val(0);}


    });

    </script>

Answer 1

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));

## EDITED ##

If you want to use it with your that format just:

Your code will look like:

 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 } $("#startdate, #enddate").change(function() { var d1 = $("#startdate").val(); var d2 = $("#enddate").val(); var minutes = 1000 * 60; var hours = minutes * 60; var day = hours * 24; var startdate1 = new Date(d1); var enddate1 = new Date(d2); var newstartdate = new Date(); newstartdate.setFullYear(startdate1.getYear(), startdate1.getMonth(), startdate1.getDay()); var newenddate = new Date(); newenddate.setFullYear(enddate1.getYear(), enddate1.getMonth(), enddate1.getDay()); var days = calcBusinessDays(newstartdate, newenddate); if (days > 0) { $("#noofdays").val(days); } else { $("#noofdays").val(0); } });

 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <label>Start Date <input type="date" id="startdate" value="2019-03-03"/> </label> <label>End Date <input type="date" id="enddate" value="2019-03-06"/> </label> <label>N. of days <output id="noofdays"/> </label>

Answer 2

To do this, you should NOT search all days between these dates !

It's not complicated, look some evident assumptions:

1. All full-week has 7-days.

2. Which 2 are weekend-days.

3. And which 5 are business-day.

Evident conclusions:

1. Look all days is loss of time.

2. Check what day is weekend to all week is loss of time.

---

Without tedious explanation.. let me show the code:

 function getBusinessDateCount (startDate, endDate) { var elapsed, daysBeforeFirstSaturday, daysAfterLastSunday; var ifThen = function (a, b, c) { return a == b ? c : a; }; elapsed = endDate - startDate; elapsed /= 86400000; daysBeforeFirstSunday = (7 - startDate.getDay()) % 7; daysAfterLastSunday = endDate.getDay(); elapsed -= (daysBeforeFirstSunday + daysAfterLastSunday); elapsed = (elapsed / 7) * 5; elapsed += ifThen(daysBeforeFirstSunday - 1, -1, 0) + ifThen(daysAfterLastSunday, 6, 5); return Math.ceil(elapsed); } function calc() { let start = document.querySelector('#startDate').value, end = document.querySelector('#endDate').value, result = getBusinessDateCount(new Date(start), new Date(end)); document.querySelector('#result').value = result; }

 Start date: <input type="date" id="startDate" value="2020-01-04"><br> End date: <input type="date" id="endDate" value="2020-01-06"><br> <input type="button" onclick="calc()" value="Get business days"><br> Business days: <input id="result" readonly>

You can test it yourself with any dates.

I just want to notice that this code ONLY consumed 0.43 sec between dates from 2000 to 2015... It is much more fast than some other codes.

Hope it helps...

Nice coding !!

Answer 3

This is how I would do it

function getDays(d1, d2) {
    var one_day=1000*60*60*24;
    var d1_days = parseInt(d1.getTime()/one_day) - 1;
    var d2_days = parseInt(d2.getTime()/one_day);
    var days = (d2_days - d1_days);
    var weeks = (d2_days - d1_days) / 7;
    var day1 = d1.getDay();
    var day2 = d2.getDay();
    if (day1 == 0) {
        days--;
    } else if (day1 == 6) {
        days-=2;
    }
    if (day2 == 0) {
       days-=2;
    } else if (day2 == 6) {
       days--;
    }
    days -= parseInt(weeks) * 2;
    alert(days);
}

getDays(new Date("June 8, 2004"),new Date("February 6, 2010"));

EDIT
To clarify my comment to @keenebec...
That solution will work for small date differences quite nicely and is easy to understand. But take something as "short" as a 6 year span and you can see a remarkable difference in speed.

http://jsfiddle.net/aSvxv/

I included all 3 answers and the original answer is indeed the fastest, but not by much and the trade off for a few microseconds of execution is somewhat trivial to me in favor of readability.

Answer 4

 Date.prototype.addDays = function(days) { var date = new Date(this.valueOf()) date.setDate(date.getDate() + days); return date; } function getBusinessDatesCount(startDate, endDate) { var count = 0; var curDate = startDate; while (curDate <= endDate) { var dayOfWeek = curDate.getDay(); var isWeekend = (dayOfWeek == 6) || (dayOfWeek == 0); if(!isWeekend) count++; curDate = curDate.addDays(1); } return count; } //Usage var startDate = new Date('7/16/2015'); var endDate = new Date('7/20/2015'); var numOfDays = getBusinessDatesCount(startDate,endDate); jQuery('div#result').text(numOfDays);

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="result"/>

Answer 5

That looks like too much work to me. I'd rather let the computer do the heavy lifting- //

Date.bizdays= function(d1, d2){
    var bd= 0, dd, incr=d1.getDate();
    while(d1<d2){
        d1.setDate(++incr);
        dd= d1.getDay();
        if(dd%6)++bd;
    }
    return bd;
}

//test

var day1= new Date(2010, 7, 11), day2= new Date(2010, 7, 31);

alert(Date.bizdays(day1, day2))

Answer 6

To understand way.,

---

1. Actual days = 14
2. weeks for Actual days = 14/7=2
3. Weekends per week=2
4. Total weekends=2*weeks for days

---

So apply this ,

 $('#EndDate').on('change', function () {
            var start = $('#StartDate').datepicker('getDate');
            var end = $('#EndDate').datepicker('getDate');
            if (start < end) {
                var days = (end - start) / 1000 / 60 / 60 / 24;

                var Weeks=Math.round(days)/7;

                var totalWeekends=Math.round(Weeks)*2;

                var puredays=Math.round(days)-totalWeekends;

                $('#days').text(Math.round(puredays) + "Working Days");


            }
            else {
alert("");
}

Thank you !

Answer 7

Important: Most answers here don't actually work if the start date (or sometimes the end date) is a saturday or sunday. I took the accepted response and modified it so that this issue is resolved now:

    var dateDiff;
    if (dateTo < dateFrom) return -1; // error code if dates transposed
    var dateFromDayOrig = dateFrom.getDay(); // day of week
    var dateToDayOrig = dateTo.getDay();
    var dateFromDay = (dateFromDayOrig == 0) ? 7 : dateFromDayOrig; // change Sunday from 0 to 7
    var dateToDay = (dateToDayOrig == 0) ? 7 : dateToDayOrig;
    dateFromDay = (dateFromDay > 5) ? 5 : dateFromDay; // only count weekdays
    dateToDay = (dateToDay > 5) ? 5 : dateToDay;

    // calculate differnece in weeks (1000mS * 60sec * 60min * 24hrs * 7 days = 604800000)
    var weekDifference = Math.floor((dateTo.getTime() - dateFrom.getTime()) / 604800000);

    if (dateFromDay <= dateToDay) {
        dateDiff = (weekDifference * 5) + (dateToDay - dateFromDay);
    } else {
        dateDiff = ((weekDifference + 1) * 5) - (dateFromDay - dateToDay);
    }

    // fix: remove one day if it's saturday or sunday
    if (dateFromDayOrig >= 6 || dateFromDayOrig == 0) {
    dateDiff--;
    }

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

Answer 8

There seems to be few issues with the response that has been marked as solution.

• The statement setFullYear() is returning incorrect value if I choose start date as 06/11/2015. So instead, the startDate1 and endDate1 can be directly passed to the function.

• If the start date is Saturday or Sunday, still the code is counting it(iWeekday1) as 5 days

• If the end date is Saturday or Sunday, still the code is counting it(iWeekday2) as 5 days. But these 5 days already get counted in the iweeks calculation.
  So instead of
  iWeekday1 = (iWeekday1 > 5) ? 5 : iWeekday1; // only count weekdays
  iWeekday2 = (iWeekday2 > 5) ? 5 : iWeekday2;
  it should be
  iWeekday1 = (iWeekday1 > 5) ? 0 : iWeekday1; // only count weekdays
  iWeekday2 = (iWeekday2 > 5) ? 0 : iWeekday2;

• The last IF condition should be executed when start and end date day is same like both are on same day, the date could be different
  if (iWeekday1 < = iWeekday2)

• The condition that adjusts if both days are weekends can be removed
  iDateDiff -= iAdjust

• Lastly, the +1 should be done only if start and end date falls on weekdays. Currently, it is adding in both the cases.
  return (iDateDiff + 1); //Add condition to apply only if both days are weekdays

--can't comment on that answer as I do not have that reputation :)

Answer 9

What I did

function calcbusinessdays()
{
    for(var c=0,e=0,d=new Date($("#startdate").val()),a=(new Date($("#enddate").val())-d)/864E5;0<=a;a--)
    {
        var b=new Date(d);
        b.setDate(b.getDate()+a);
        1==Math.ceil(b.getDay()%6/6)?c++:e++
    }
    $("#noofdays").html(c)
};

c is weekdays, e is weekends

Answer 10

I get it work with this code. Note that the function is from date.js and businessday js (thanks to Garis Suero). Start Date 11-08-2010 End Date 16-08-2010 will result 4 days of leave.

<script type="text/javascript">

    $("#startdate, #enddate").change(function() {       

    var d1 = $("#startdate").val();
    var d2 = $("#enddate").val();

            var minutes = 1000*60;
            var hours = minutes*60;
            var day = hours*24;

            var startdate1 = getDateFromFormat(d1, "d-m-y");
            var enddate1 = getDateFromFormat(d2, "d-m-y");

            var days = calcBusinessDays(new Date(startdate1),new Date(enddate1));             

    if(days>0)
    { $("#noofdays").val(days);}
    else
    { $("#noofdays").val(0);}


    });

    </script>

Answer 11

function addDays(date, days) {
    var result = new Date(date);
    result.setDate(result.getDate() + days);
    return result;
}
 var currentDate;
                            selectFlixbleDates = [];
                            var monToSatDateFilter=[];
                            currentDate=new Date(date);
                            while(currentDate){
                                console.log("currentDate"+currentDate);
                                if(new Date(currentDate).getDay()!=0){
                                    selectFlixbleDates.push(currentDate)
                                }
                                if(selectFlixbleDates.length==$scope.numberOfDatePick)
                                {

                                    break;
                                }
                                currentDate=addDays(currentDate,1);

                            }
                            for (var i = 0; i < selectFlixbleDates.length; i++) {

                                //  console.log(between[i]);
                                monToSatDateFilter.push((selectFlixbleDates[i].getMonth() + 1) + '/' + selectFlixbleDates[i].getDate() + '/' + selectFlixbleDates[i].getFullYear());

                            }
                            var endDate=monToSatDateFilter.slice(-1).pop();
                            var space =monToSatDateFilter.join(', ');
                            var sdfs= document.getElementById("maxPicks").value =space;
                            $scope.$apply(function() {
                                $scope.orderEndDate=monToSatDateFilter.slice(-1).pop()
                                $scope.orderStartDate=monToSatDateFilter[0];
                            });
                            document.getElementById("startDateEndDate").innerHTML =$scope.orderStartDate+ ' TO ' +$scope.orderEndDate
                        }  

Answer 12

I have used Angular framework and Moment.js library to implement the solution. My solution covers all the cases.

this.daysInBetween = this.endMoment.diff(this.startMoment, 'days') + 1;
this.weeksInBetween = this.endMoment.diff(this.startMoment, 'weeks');
this.weekDays = this.daysInBetween - (this.weeksInBetween * 2);

if ( (this.startMoment.day() === 0 && this.endMoment.day() === 6) ||
 (this.startMoment.day() > this.endMoment.day()) ) {
  // IF ONE WEEKEND WAS MISSED
  this.weekDays-=2;
} else if ( this.startMoment.day() <= this.endMoment.day() &&
( this.startMoment.day() === 0 || this.startMoment.day() === 6 ||
  this.endMoment.day() === 0 || this.endMoment.day() === 6) ) {
  // IF EITHER OF DAYS WAS A WEEKEND
  this.weekDays--;
}

Live Demo: Calculate number of weekdays

I am currently working on a blog to write about my approach to this specific problem. I will post the link to the blog on the comment.

Answer 13

Important: Most answers here don't actually work if the start date (or sometimes the end date) is a saturday or sunday. For example: if your start and end date are

• Saturday to Sunday or vice versa
• or Saturday to Saturday
• or Sunday to Sunday

So, here is the modified answer from the accepted answer

function calculateBusinessDays(dDate1, dDate2) {

var iWeeks, iDateDiff, iAdjust = 0;
if (dDate2 < dDate1) {
    return -1;
}

var iWeekday1 = dDate1.getDay();
var iWeekday2 = dDate2.getDay();

iWeekday1 = (iWeekday1 === 0) ? 7 : iWeekday1;
iWeekday2 = (iWeekday2 === 0) ? 7 : iWeekday2;

if (iWeekday1 > 5 && iWeekday2 <= 6) {
    iWeekday1 = 0;
    iAdjust = 1;
}
iWeekday2 = (iWeekday1 === 0 && iWeekday2 === 6) ? 0 : iWeekday2;

if ((iWeekday1 > 5) && (iWeekday2 > 5)) iAdjust = 1;

iWeekday1 = (iWeekday1 > 5) ? 5 : iWeekday1;
iWeekday2 = (iWeekday2 > 5) ? 5 : iWeekday2;
iWeeks = Math.floor((dDate2.getTime() - dDate1.getTime()) / 604800000);

if (iWeeks===0 && iWeekday1===0 && iWeekday2===0 
    && (dDate2.getTime() !== dDate1.getTime()) ) {
    iWeeks = 1;
}

if (iWeekday1 <= iWeekday2) {
    iDateDiff = (iWeeks * 5) + (iWeekday2 - iWeekday1)
} else {
    iDateDiff = ((iWeeks + 1) * 5) - (iWeekday1 - iWeekday2)
}
iDateDiff -= iAdjust
return (iDateDiff + 1);

}