查找两个日期之间的天差(不包括周末) [英] Find day difference between two dates (excluding weekend days)

问题描述

我正在使用 jquery-ui datepicker 来选择日期和 date.js找出 2 个日期之间的差异.

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

现在的问题是我想从计算中排除周末(周六和周日).我该怎么做?

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

例如，用户选择开始日期 (13/8/2010) 和结束日期 (16/8/2010).由于 14/8/2010 和 15/8/2010 是在工作日，而不是总共 4 天，我希望它只有 2 天.

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.

这是我现在使用的代码:

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>

推荐答案

也许其他人可以帮你把这个功能转换成 JQuery 的框架...

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

我在这里找到了这个功能.

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));

如果你想以你的格式使用它:

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

您的代码将如下所示:

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>

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