查找两个日期之间的日差（不包括周末）

一尘不染

查找两个日期之间的日差（不包括周末）

javascript

嗨，我正在使用jquery-uidatepicker选择date和date.js来查找两个日期之间的差异。

现在的问题是我想从计算中排除周末（周六和周日）。我该怎么办？

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是工作日，而不是总共4天，所以我希望只有2天。

这是即时消息正在使用的代码：

<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>

阅读 317

2020-05-01

共1个答案

一尘不染

也许其他人可以帮助您将此功能转换为JQuery的框架…

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));

##编辑

如果您想将其与该格式一起使用，请执行以下操作：

您的代码如下所示：

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>

2020-05-01