Javascript中日期范围内的特定天数

Question

Javascript中日期范围内的特定天数

use*_*647 2 javascript timespan dayofweek

我有两个约会.一个是开始日期,另一个是结束日期.我想计算星期六,星期一和星期三在日期范围内的数量？我该如何解决？我看了几个教程,但他们只计算日期范围内的日期.提前致谢.我使用以下代码来计算工作日,但我只需要在星期六,星期一和星期三的日期范围内.

<!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>

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Answer 1

Mat*_*att 7

O(1)解决方案.一周中的天数(不超过7天),而不是日期范围.

// 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);
}

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例如,计算2014年1月的星期一(1),星期三(3)和星期六(6):

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

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注意:这假定d0并且d1是Date时间大致相同的对象.如果创建的Date对象仅指定年,月和日,则没有问题.

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