All geek questions in one place

How many specific days in a date range in Javascript

I have two dates. The first is the start date, and the other is the end date. I want to calculate how many Saturdays, Mondays and Wednesdays are in a date range? How can i solve this? I have seen a few lessons, but they only consider dates in a date range. Thank you in advance. I use the following code to calculate only working days, but I only need how many Saturn, Mondays and Tuesdays are in the date range.

<!DOCTYPE html>
<html>
<head>
<meta charset=utf-8 />
<title>JS Bin</title>
<!--[if IE]>
  <script src="http://html5shiv.googlecode.com/svn/trunk/html5.js"></script>
<![endif]-->
  <script>
        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) {
          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
    }
    </script>
<style>
  article, aside, figure, footer, header, hgroup, 
  menu, nav, section { display: block; }
</style>
</head>
<body>
  <script>
    alert(calcBusinessDays(new Date("August 01, 2010 11:13:00"),new Date("August 31, 2010 11:13:00")));
  </script>
</body>
</html>
+1
source share
2 answers

My approach:

First get a list of dates from the range between two dates:

function getDates(dateStart, dateEnd) {
  var currentDate = dateStart,
      dates = [];
  while(currentDate <= dateEnd) {

    // append date to array
    dates.push(currentDate);

    // add one day
    // automatically rolling over to next month
    var d = new Date(currentDate.valueOf());
    d.setDate(d.getDate() + 1);
    currentDate = d;

  }
  return dates;
}

Then cyclically repeat these dates, filtering for the corresponding workday index:

function filterWeekDays(dates, includeDays) {
  var weekdays = [];

  // cycle dates
  dates.forEach(function(day){

    // cycle days to be included (so==0, mo==1, etc.)
    includeDays.forEach(function(include) {
      if(day.getDay() == include) {
        weekdays.push(day);
      }
    });
  });
  return weekdays;
}

:

, 0 1, Sun, Mon, Wed : [0, 1, 3]

JS Bin: http://jsbin.com/furupoqu/5/edit

+2

O (1). ( 7), .

// days is an array of weekdays: 0 is Sunday, ..., 6 is Saturday
function countCertainDays( days, d0, d1 ) {
  var ndays = 1 + Math.round((d1-d0)/(24*3600*1000));
  var sum = function(a,b) {
    return a + Math.floor( (ndays+(d0.getDay()+6-b) % 7 ) / 7 ); };
  return days.reduce(sum,0);
}

, (1), (3) (6) 2014 :

countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,1)) // 1
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,2)) // 1
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,3)) // 1
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,4)) // 2
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,5)) // 2
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,6)) // 3
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,7)) // 3
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,8)) // 4
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,9)) // 4
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,10)) // 4
countCertainDays([1,3,6],new Date(2014,0,1),new Date(2014,0,11)) // 5

. , d0 d1 Date, . Date, , , .

+3

Source: https://habr.com/ru/post/1606074/

More articles:


All Articles