刚遇到这个问题，最简单的方法就是用moment:

你需要先安装moment和moment-range:

const Moment = require('moment');
const MomentRange = require('moment-range');
const moment = MomentRange.extendMoment(Moment);

const start = moment()
const end = moment().add(2, 'months')
const range = moment.range(start, end)
const arrayOfDates = Array.from(range.by('days'))
console.log(arrayOfDates)

var boxingDay = new Date("12/26/2010");
var nextWeek  = boxingDay*1 + 7*24*3600*1000;

function getDates( d1, d2 ){
  var oneDay = 24*3600*1000;
  for (var d=[],ms=d1*1,last=d2*1;ms<last;ms+=oneDay){
    d.push( new Date(ms) );
  }
  return d;
}

getDates( boxingDay, nextWeek ).join("\n");
// Sun Dec 26 2010 00:00:00 GMT-0700 (Mountain Standard Time)
// Mon Dec 27 2010 00:00:00 GMT-0700 (Mountain Standard Time)
// Tue Dec 28 2010 00:00:00 GMT-0700 (Mountain Standard Time)
// Wed Dec 29 2010 00:00:00 GMT-0700 (Mountain Standard Time)
// Thu Dec 30 2010 00:00:00 GMT-0700 (Mountain Standard Time)
// Fri Dec 31 2010 00:00:00 GMT-0700 (Mountain Standard Time)
// Sat Jan 01 2011 00:00:00 GMT-0700 (Mountain Standard Time)

你可以使用momentJS轻松做到这一点

给你的依赖增加时间

npm i moment

然后把它导入你的文件

var moment = require("moment");

然后使用下面的代码获取两个日期之间的所有日期的列表

let dates = [];
let currDate = moment.utc(new Date("06/30/2019")).startOf("day");
let lastDate = moment.utc(new Date("07/30/2019")).startOf("day");

do {
 dates.push(currDate.clone().toDate());
} while (currDate.add(1, "days").diff(lastDate) < 0);
dates.push(currDate.clone().toDate());

console.log(dates);

不是最短的，而是简单的，不可变的，没有依赖关系

function datesArray(start, end) {
    let result = [], current = new Date(start);
    while (current <= end)
        result.push(current) && (current = new Date(current)) && current.setDate(current.getDate() + 1);
    return result;
}

使用

函数datesArray(start, end) { let result = []， current = new Date(start); While (current <= end) result.push(current) && (current = new Date(current)) && current. setdate (current. getdate () + 1); 返回结果; } / /使用 const test = datesArray( 新的日期(“2020-02-26”), 新日期(“2020-03-05”) ）; 对于(设I = 0;I < test.length;I ++) { console.log ( 测试[我].toISOString () .slice (0, 10) ）; }

Date.prototype.addDays = function(days) {
    var date = new Date(this.valueOf());
    date.setDate(date.getDate() + days);
    return date;
}

function getDates(startDate, stopDate) {
    var dateArray = new Array();
    var currentDate = startDate;
    while (currentDate <= stopDate) {
        dateArray.push(new Date (currentDate));
        currentDate = currentDate.addDays(1);
    }
    return dateArray;
}

这里是一个功能演示http://jsfiddle.net/jfhartsock/cM3ZU/

我看了上面所有的。最后我自己开始写作。你不需要花时间。原生的for循环就足够了，而且最有意义，因为for循环的存在是为了计算一个范围内的值。

一个衬套:

var getDaysArray = function(s,e) {for(var a=[],d=new Date(s);d<=new Date(e);d.setDate(d.getDate()+1)){ a.push(new Date(d));}return a;};

长版本

var getDaysArray = function(start, end) {
    for(var arr=[],dt=new Date(start); dt<=new Date(end); dt.setDate(dt.getDate()+1)){
        arr.push(new Date(dt));
    }
    return arr;
};

列出中间的日期:

var daylist = getDaysArray(new Date("2018-05-01"),new Date("2018-07-01"));
daylist.map((v)=>v.toISOString().slice(0,10)).join("")
/*
Output: 
    "2018-05-01
    2018-05-02
    2018-05-03
    ...
    2018-06-30
    2018-07-01"
*/

从过去日期到现在的日子:

var daylist = getDaysArray(new Date("2018-05-01"),new Date());
daylist.map((v)=>v.toISOString().slice(0,10)).join("")