如何计算两个日期之间的天数？ [重复]

7 回答

• 14

  var oneDay = 24*60*60*1000; // hours*minutes*seconds*milliseconds
  var firstDate = new Date(2008,01,12);
  var secondDate = new Date(2008,01,22);
  
  var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
  

  回复于 2024-04-20T08:59:38+08:00
• 430

  这是一个执行此操作的函数：

  function days_between(date1, date2) {
  
      // The number of milliseconds in one day
      var ONE_DAY = 1000 * 60 * 60 * 24;
  
      // Convert both dates to milliseconds
      var date1_ms = date1.getTime();
      var date2_ms = date2.getTime();
  
      // Calculate the difference in milliseconds
      var difference_ms = Math.abs(date1_ms - date2_ms);
  
      // Convert back to days and return
      return Math.round(difference_ms/ONE_DAY);
  
  }
  

  回复于 2024-04-20T08:59:38+08:00
• 0

  这是我使用的 . 如果您只是减去日期，它将无法在夏令时间界限内工作（例如4月1日至4月30日或10月1日至10月31日） . 这会减少所有时间，以确保您获得一天，并通过使用UTC消除任何DST问题 .

  var nDays = (    Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate()) -
                   Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate())) / 86400000;
  

  回复于 2024-04-20T08:59:38+08:00
• 9

  这是我的实现：

  function daysBetween(one, another) {
    return Math.round(Math.abs((+one) - (+another))/8.64e7);
  }
  

  +<date> 对整数表示执行类型强制，并且与 <date>.getTime() 具有相同的效果， 8.64e7 是一天中的毫秒数 .

  回复于 2024-04-20T08:59:38+08:00
• 8

  调整以允许夏令时差异 . 试试这个：

  function daysBetween(date1, date2) {
  
   // adjust diff for for daylight savings
   var hoursToAdjust = Math.abs(date1.getTimezoneOffset() /60) - Math.abs(date2.getTimezoneOffset() /60);
   // apply the tz offset
   date2.addHours(hoursToAdjust); 
  
      // The number of milliseconds in one day
      var ONE_DAY = 1000 * 60 * 60 * 24
  
      // Convert both dates to milliseconds
      var date1_ms = date1.getTime()
      var date2_ms = date2.getTime()
  
      // Calculate the difference in milliseconds
      var difference_ms = Math.abs(date1_ms - date2_ms)
  
      // Convert back to days and return
      return Math.round(difference_ms/ONE_DAY)
  
  }
  
  // you'll want this addHours function too 
  
  Date.prototype.addHours= function(h){
      this.setHours(this.getHours()+h);
      return this;
  }
  

  回复于 2024-04-20T08:59:38+08:00
• 7

  我已经为另一个帖子写了这个解决方案，他们问，如何计算两个日期之间的差异，所以我分享了我准备的内容：

  // Here are the two dates to compare
  var date1 = '2011-12-24';
  var date2 = '2012-01-01';
  
  // First we split the values to arrays date1[0] is the year, [1] the month and [2] the day
  date1 = date1.split('-');
  date2 = date2.split('-');
  
  // Now we convert the array to a Date object, which has several helpful methods
  date1 = new Date(date1[0], date1[1], date1[2]);
  date2 = new Date(date2[0], date2[1], date2[2]);
  
  // We use the getTime() method and get the unixtime (in milliseconds, but we want seconds, therefore we divide it through 1000)
  date1_unixtime = parseInt(date1.getTime() / 1000);
  date2_unixtime = parseInt(date2.getTime() / 1000);
  
  // This is the calculated difference in seconds
  var timeDifference = date2_unixtime - date1_unixtime;
  
  // in Hours
  var timeDifferenceInHours = timeDifference / 60 / 60;
  
  // and finaly, in days :)
  var timeDifferenceInDays = timeDifferenceInHours  / 24;
  
  alert(timeDifferenceInDays);
  

  您可以跳过代码中的一些步骤，我已编写它以便于理解 .

  你会在这里找到一个正在运行的例子：http://jsfiddle.net/matKX/

  回复于 2024-04-20T08:59:38+08:00
• 32

  从我的小日期差异计算器：

  var startDate = new Date(2000, 1-1, 1);  // 2000-01-01
  var endDate =   new Date();              // Today
  
  // Calculate the difference of two dates in total days
  function diffDays(d1, d2)
  {
    var ndays;
    var tv1 = d1.valueOf();  // msec since 1970
    var tv2 = d2.valueOf();
  
    ndays = (tv2 - tv1) / 1000 / 86400;
    ndays = Math.round(ndays - 0.5);
    return ndays;
  }
  

  所以你会打电话给：

  var nDays = diffDays(startDate, endDate);
  

  （完整来源于http://david.tribble.com/src/javascript/jstimespan.html . ）

  Addendum

  可以通过更改这些行来改进代码：

  var tv1 = d1.getTime();  // msec since 1970
    var tv2 = d2.getTime();
  

  回复于 2024-04-20T08:59:38+08:00