var range = getDates(new Date(), new Date().addDays(7));

我想“范围”是一个日期对象的数组,一个为两个日期之间的每一天。

诀窍在于它还应该处理月份和年份的边界。


当前回答

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) ); }

其他回答

getDates = (from, to) => {
    const cFrom = new Date(from);
    const cTo = new Date(to);

    let daysArr = [new Date(cFrom)];
    let tempDate = cFrom;

    while (tempDate < cTo) {
        tempDate.setUTCDate(tempDate.getUTCDate() + 1);
        daysArr.push(new Date(tempDate));
    }

    return daysArr;
}
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/

这可能会帮助到一些人,

您可以从中获得行输出,并根据需要格式化row_date对象。

var from_date = '2016-01-01';
var to_date = '2016-02-20';

var dates = getDates(from_date, to_date);

console.log(dates);

function getDates(from_date, to_date) {
  var current_date = new Date(from_date);
  var end_date     = new Date(to_date);

  var getTimeDiff = Math.abs(current_date.getTime() - end_date.getTime());
  var date_range = Math.ceil(getTimeDiff / (1000 * 3600 * 24)) + 1 ;

  var weekday = ["SUN", "MON", "TUE", "WED", "THU", "FRI", "SAT"];
  var months = ["JAN", "FEB", "MAR", "APR", "MAY", "JUN", "JUL", "AUG", "SEP", "OCT", "NOV", "DEC"];
  var dates = new Array();

  for (var i = 0; i <= date_range; i++) {
     var getDate, getMonth = '';

     if(current_date.getDate() < 10) { getDate = ('0'+ current_date.getDate());}
     else{getDate = current_date.getDate();}

    if(current_date.getMonth() < 9) { getMonth = ('0'+ (current_date.getMonth()+1));}
    else{getMonth = current_date.getMonth();}

    var row_date = {day: getDate, month: getMonth, year: current_date.getFullYear()};
    var fmt_date = {weekDay: weekday[current_date.getDay()], date: getDate, month: months[current_date.getMonth()]};
    var is_weekend = false;
    if (current_date.getDay() == 0 || current_date.getDay() == 6) {
        is_weekend = true;
    }
    dates.push({row_date: row_date, fmt_date: fmt_date, is_weekend: is_weekend});
    current_date.setDate(current_date.getDate() + 1);
 }
 return dates;
}

https://gist.github.com/pranid/3c78f36253cbbc6a41a859c5d718f362.js

Generate an array of years: const DAYS = () => { const days = [] const dateStart = moment() const dateEnd = moment().add(30, ‘days') while (dateEnd.diff(dateStart, ‘days') >= 0) { days.push(dateStart.format(‘D')) dateStart.add(1, ‘days') } return days } console.log(DAYS()) Generate an arrays for month: const MONTHS = () => { const months = [] const dateStart = moment() const dateEnd = moment().add(12, ‘month') while (dateEnd.diff(dateStart, ‘months') >= 0) { months.push(dateStart.format(‘M')) dateStart.add(1, ‘month') } return months } console.log(MONTHS()) Generate an arrays for days: const DAYS = () => { const days = [] const dateStart = moment() const dateEnd = moment().add(30, ‘days') while (dateEnd.diff(dateStart, ‘days') >= 0) { days.push(dateStart.format(‘D')) dateStart.add(1, ‘days') } return days } console.log(DAYS())

这里有一个不需要任何库的代码行,以防你不想创建另一个函数。只需用变量或日期值替换startDate(在两个地方)和endDate(这是js的日期对象)。当然,如果你愿意,你可以把它包装在一个函数中

Array(Math.floor((endDate - startDate) / 86400000) + 1).fill().map((_, idx) => (new Date(startDate.getTime() + idx * 86400000)))