# 如何计算两个日期之间的天数？[重复]

2020/10/06 18:01 · javascript ·  · 0评论
1. 我正在计算从“开始”到“结束”日期之间的天数。例如，如果起始日期为2010年4月13日，起始日期为2010年5月15日，则结果应为

2. 如何使用JavaScript获得结果？

``````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));
``````

``````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);

}
``````

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

``````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;
}
``````

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

`+<date>`将类型强制转换为整数表示形式，并且具有与相同的效果，`<date>.getTime()`并且`8.64e7`是一天中的毫秒数。

``````  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;
}
``````

``````// 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);
``````

``````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;
}
``````

``````var nDays = diffDays(startDate, endDate);
``````

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