如何计算两个日期之间的天数？ [复制]

Question

1. 我正在计算“开始”和“结束”日期之间的天数。例如，如果起始日期是 13/04/2010，截止日期是 15/04/2010，那么结果应该是

2. 如何使用 JavaScript 获得结果？

Answer 1

这是一个执行此操作的函数：

function days_between(date1, date2) {

    // The number of milliseconds in one day
    const ONE_DAY = 1000 * 60 * 60 * 24;

    // Calculate the difference in milliseconds
    const differenceMs = Math.abs(date1 - date2);

    // Convert back to days and return
    return Math.round(differenceMs / ONE_DAY);

}

Answer 2

const oneDay = 24 * 60 * 60 * 1000; // hours*minutes*seconds*milliseconds
const firstDate = new Date(2008, 1, 12);
const secondDate = new Date(2008, 1, 22);

const diffDays = Math.round(Math.abs((firstDate - secondDate) / oneDay));

Answer 3

这是我使用的。如果您只是减去日期，它将无法跨越夏令时边界（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有一天，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;

作为一个函数：

function DaysBetween(StartDate, EndDate) {
  // The number of milliseconds in all UTC days (no DST)
  const oneDay = 1000 * 60 * 60 * 24;

  // A day in UTC always lasts 24 hours (unlike in other time formats)
  const start = Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate());
  const end = Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate());

  // so it's safe to divide by 24 hours
  return (start - end) / oneDay;
}

Answer 4

来自我的小日期差计算器：

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：

var nDays = diffDays(startDate, endDate);

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

附录

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();

Answer 5

这是我的实现：

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

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

Answer 6

我已经为另一个询问如何计算两个日期之间的差异的帖子编写了这个解决方案，所以我分享我准备的内容：

// 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);

你可以跳过代码中的一些步骤，我写了它以便于理解。

您将在此处找到一个运行示例：http://jsfiddle.net/matKX/

Answer 7

进行了调整以考虑夏令时差异。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;
}