Question

Hi i正在使用jquery-uidatepicker来选定日期和date.js,以便在两个日期之间找到区别。

现在的问题是,我想将周末的计算时间(日间和日间)排除在外。如何这样做?

例如,用户选择起始日期(2010年8月13日)和截止日期(2010年8月16日)。自2006年以来 2010年8月14日和2010年8月15日是每周的,而不是总共4天,只需要2天。

这是现在使用的法典:

<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

也许其他人可以帮助你将这一职能转换成宗教基金框架......

我发现这一职能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 ##

如果你想用你这样的格式使用:

你的法典将考虑:

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

为此,你请NOT在这些日期之间搜索所有天!

它并不复杂,只看一些明显的假设:

所有全周都有7天。
2个周末。
其中5个为营业日。

证据确凿的结论:

一切日子都是浪费时间。
周末到所有一周检查是时间损失。

如果没有令人不解的解释,请让我表明:

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>

你们可以用任何日期来测试它。

我只想指出,,该法典在2000年至2015年期间消费了0.43 sec 。这比其他一些法典要快得多。

希望有助于......

Nice!

Answer 3

这就是我如何这样做。

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/

我包括所有3个答案,最初的答案确实是最快的,但不是很多答案,对几个微秒的处决进行交易对我来说有些微不足道,有利于阅读。

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

理解方式,

Actual days = 14
weeks for Actual days = 14/7=2
Weekends per week=2
Total weekends=2*weeks for days

因此,

 $( #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("");
}

谢谢!

Answer 7

<>Important: 如果启动日期(有时是结束日期)为饱和日,这里的大多数答案实际上不会奏效。我采取了已接受的答复,并作了修改,以便现在解决这个问题:

    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

看来,作为解决办法的反应很少。

如果我选择起算日期为2015年11月6日,则该表将“未来年”重新计算。因此,起始点1和最终结果1可直接转至这一职能。
如果起算日期是星期六或星期天,则该守则将它(i Weekday1)计算为5天。
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

--------------------------

Answer 9

我开始这项工作。请注意,该职能来自date.js和营业日js(与Gardis Suero相比)。开始日期 * E/CN.6/2009/1。 16-08-2010年将有4天的休假。

<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 10

我做了些什么。

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)
};

<代码>c为周日,e为周末。

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

因此,这里是从已接受的答复中修改的答案。

function calculateBusinessDays(dDate1, dDate2) {

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

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;
iii
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;
iii

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

iii

## EDITED ##

友情链接