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

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;
}

其他回答

@softvar的解决方案,但随后包括工作日期选项

/**
 * Returns array of working days between two dates.
 *
 * @param {string} startDate
 *   The start date in yyyy-mm-dd format.
 * @param {string} endDate
 *   The end date in yyyy-mm-dd format.
 * @param {boolean} onlyWorkingDays
 *   If true only working days are returned. Default: false
 *
 * @return {array}
 *   Array of dates in yyyy-mm-dd string format.
 */
function getDates(startDate, stopDate, onlyWorkingDays) {
  let doWd = typeof onlyWorkingDays ==='undefined' ? false : onlyWorkingDays;

  let dateArray = [];  
  let dayNr;
  let runDateObj = moment(startDate);  
  let stopDateObj = moment(stopDate);

  while (runDateObj <= stopDateObj) {
    dayNr = runDateObj.day();
    if (!doWd || (dayNr>0 && dayNr<6)) {
     dateArray.push(moment(runDateObj).format('YYYY-MM-DD'));  
    }

    runDateObj = moment(runDateObj).add(1, 'days');
  }
  return dateArray;
}
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)

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) ); }

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);

功能:

  var dates = [],
      currentDate = startDate,
      addDays = function(days) {
        var date = new Date(this.valueOf());
        date.setDate(date.getDate() + days);
        return date;
      };
  while (currentDate <= endDate) {
    dates.push(currentDate);
    currentDate = addDays.call(currentDate, 1);
  }
  return dates;
};

用法:

var dates = getDatesRange(new Date(2019,01,01), new Date(2019,01,25));                                                                                                           
dates.forEach(function(date) {
  console.log(date);
});

希望对你有所帮助