例如,在输入框中给定两个日期:

<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>

<script>
  alert(datediff("day", first, second)); // what goes here?
</script>

如何在JavaScript中获得两个日期之间的天数?


当前回答

下面的解决方案将假设这些变量在代码中可用:

const startDate  = '2020-01-01';
const endDate    = '2020-03-15';

原生 JS

步骤:

设定开始日期 设定结束日期 计算的区别 将毫秒转换为天

const diffInMs   = new Date(endDate) - new Date(startDate)
const diffInDays = diffInMs / (1000 * 60 * 60 * 24);

备注:

我知道这不是你问题的一部分,但一般来说,我不建议在香草JavaScript中做任何日期计算或操作,而是使用date-fns, Luxon或moment.js这样的库,因为有许多边缘情况。

这个简单的JavaScript回答以十进制数计算天数。此外,在使用夏令时时,它可能会遇到边缘情况


使用图书馆

- 日期-fns

const differenceInDays = require('date-fns/differenceInDays');
const diffInDays = differenceInDays(new Date(endDate), new Date(startDate));

文档:https://date-fns.org/v2.16.1/docs/differenceInDays

——国际光子

const { DateTime } = require('luxon');
const diffInDays = DateTime.fromISO(endDate).diff(DateTime.fromISO(startDate), 'days').toObject().days;

文档:https://moment.github.io/luxon/docs/class/src/datetime.js DateTime.html # instance-method-diff

——Moment.js

const moment = require('moment');
const diffInDays = moment(endDate).diff(moment(startDate), 'days');

文档:https://momentjs.com/docs/ / /显示/不同


RunKit示例

其他回答

计算两个日期之间天数的简单方法是去掉它们的时间分量,即将小时、分钟、秒和毫秒设置为0,然后减去它们的时间,用一天的毫秒值来计算。

var firstDate= new Date(firstDate.setHours(0,0,0,0));
var secondDate= new Date(secondDate.setHours(0,0,0,0));
var timeDiff = firstDate.getTime() - secondDate.getTime();
var diffDays =timeDiff / (1000 * 3600 * 24);

如果你想有一个DateArray日期试试这个:

<script>
        function getDates(startDate, stopDate) {
        var dateArray = new Array();
        var currentDate = moment(startDate);
        dateArray.push( moment(currentDate).format('L'));

        var stopDate = moment(stopDate);
        while (dateArray[dateArray.length -1] != stopDate._i) {
            dateArray.push( moment(currentDate).format('L'));
            currentDate = moment(currentDate).add(1, 'days');
        }
        return dateArray;
      }
</script>

调试片段

在这种情况下使用moment会容易得多,你可以试试这个:

    let days = moment(yourFirstDateString).diff(moment(yourSecondDateString), 'days');

它会给你一个整数值,比如1、2、5、0等,所以你可以很容易地使用条件检查,比如:

if(days < 1) {

此外,还有一件事是你可以得到更准确的时间差结果(以小数形式,如1.2,1.5,0.7等),使用以下语法得到这种结果:

let days = moment(yourFirstDateString).diff(moment(yourSecondDateString), 'days', true);

如果你有任何进一步的疑问,请告诉我

我从其他答案中得到一些灵感,使输入具有自动卫生。我希望这是对其他答案的改进。

我还推荐使用<input type="date">字段,这将有助于验证用户输入。

//use best practices by labeling your constants. let MS_PER_SEC = 1000 , SEC_PER_HR = 60 * 60 , HR_PER_DAY = 24 , MS_PER_DAY = MS_PER_SEC * SEC_PER_HR * HR_PER_DAY ; //let's assume we get Date objects as arguments, otherwise return 0. function dateDiffInDays(date1Time, date2Time) { if (!date1Time || !date2Time) return 0; return Math.round((date2Time - date1Time) / MS_PER_DAY); } function getUTCTime(dateStr) { const date = new Date(dateStr); // If use 'Date.getTime()' it doesn't compute the right amount of days // if there is a 'day saving time' change between dates return Date.UTC(date.getFullYear(), date.getMonth(), date.getDate()); } function calcInputs() { let date1 = document.getElementById("date1") , date2 = document.getElementById("date2") , resultSpan = document.getElementById("result") ; if (date1.value && date2.value && resultSpan) { //remove non-date characters console.log(getUTCTime(date1.value)); let date1Time = getUTCTime(date1.value) , date2Time = getUTCTime(date2.value) , result = dateDiffInDays(date1Time, date2Time) ; resultSpan.innerHTML = result + " days"; } } window.onload = function() { calcInputs(); }; //some code examples console.log(dateDiffInDays(new Date("1/15/2019"), new Date("1/30/2019"))); console.log(dateDiffInDays(new Date("1/15/2019"), new Date("2/30/2019"))); console.log(dateDiffInDays(new Date("1/15/2000"), new Date("1/15/2019"))); <input type="date" id="date1" value="2000-01-01" onchange="calcInputs();" /> <input type="date" id="date2" value="2022-01-01" onchange="calcInputs();"/> Result: <span id="result"></span>

1970-01-01之前和2038-01-19之后的贡献

function DateDiff(aDate1, aDate2) {
  let dDay = 0;
  this.isBissexto = (aYear) => {
    return (aYear % 4 == 0 && aYear % 100 != 0) || (aYear % 400 == 0);
  };
  this.getDayOfYear = (aDate) => {
    let count = 0;
    for (let m = 0; m < aDate.getUTCMonth(); m++) {
      count += m == 1 ? this.isBissexto(aDate.getUTCFullYear()) ? 29 : 28 : /(3|5|8|10)/.test(m) ? 30 : 31;
    }
    count += aDate.getUTCDate();
    return count;
  };
  this.toDays = () => {
    return dDay;
  };
  (() => {
    let startDate = aDate1.getTime() <= aDate2.getTime() ? new Date(aDate1.toISOString()) : new Date(aDate2.toISOString());
    let endDate = aDate1.getTime() <= aDate2.getTime() ? new Date(aDate2.toISOString()) : new Date(aDate1.toISOString());
    while (startDate.getUTCFullYear() != endDate.getUTCFullYear()) {
      dDay += (this.isBissexto(startDate.getFullYear())? 366 : 365) - this.getDayOfYear(startDate) + 1;
      startDate = new Date(startDate.getUTCFullYear()+1, 0, 1);
    }
    dDay += this.getDayOfYear(endDate) - this.getDayOfYear(startDate);
  })();
}