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

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);

其他回答

   function validateDate() {
        // get dates from input fields
        var startDate = $("#startDate").val();
        var endDate = $("#endDate").val();
        var sdate = startDate.split("-");
        var edate = endDate.split("-");
        var diffd = (edate[2] - sdate[2]) + 1;
        var leap = [ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
        var nonleap = [ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
        if (sdate[0] > edate[0]) {
            alert("Please enter End Date Year greater than Start Date Year");
            document.getElementById("endDate").value = "";
            diffd = "";
        } else if (sdate[1] > edate[1]) {
            alert("Please enter End Date month greater than Start Date month");
            document.getElementById("endDate").value = "";
            diffd = "";
        } else if (sdate[2] > edate[2]) {
            alert("Please enter End Date greater than Start Date");
            document.getElementById("endDate").value = "";
            diffd = "";
        } else {
            if (sdate[0] / 4 == 0) {
                while (sdate[1] < edate[1]) {
                    diffd = diffd + leap[sdate[1]++];
                }
            } else {
                while (sdate[1] < edate[1]) {
                    diffd = diffd + nonleap[sdate[1]++];
                }
            }
            document.getElementById("numberOfDays").value = diffd;
        }
    }

在撰写本文时,其他答案中只有一个正确处理DST(夏令时)转换。以下是位于加州的一个系统的结果:

                                        1/1/2013- 3/10/2013- 11/3/2013-
User       Formula                      2/1/2013  3/11/2013  11/4/2013  Result
---------  ---------------------------  --------  ---------  ---------  ---------
Miles                   (d2 - d1) / N   31        0.9583333  1.0416666  Incorrect
some         Math.floor((d2 - d1) / N)  31        0          1          Incorrect
fuentesjr    Math.round((d2 - d1) / N)  31        1          1          Correct
toloco     Math.ceiling((d2 - d1) / N)  31        1          2          Incorrect

N = 86400000

虽然数学。round返回正确的结果,我认为它有点笨拙。相反,当DST开始或结束时,通过显式计算UTC偏移量的变化,我们可以使用精确的算术:

function treatAsUTC(date) {
    var result = new Date(date);
    result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
    return result;
}

function daysBetween(startDate, endDate) {
    var millisecondsPerDay = 24 * 60 * 60 * 1000;
    return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}

alert(daysBetween($('#first').val(), $('#second').val()));

解释

JavaScript的日期计算很棘手,因为date对象内部存储的时间是UTC,而不是本地时间。例如,3/10/2013太平洋标准时间12:00 AM (UTC-08:00)存储为3/10/2013上午8:00 UTC, 3/11/2013太平洋夏令时12:00 AM (UTC-07:00)存储为3/11/2013上午7:00 UTC。在这一天,从午夜到午夜,当地时间在UTC只有23小时!

虽然本地时间中的一天可以大于或小于24小时,但国际标准时间中的一天总是24小时上面所示的daysBetween方法利用了这一事实,它首先调用treatAsUTC将本地时间调整为午夜UTC,然后再进行减法和除法。

1. JavaScript忽略闰秒。

我认为解决方案不是100%正确的,我会使用天花板而不是地板,圆形将工作,但这不是正确的操作。

function dateDiff(str1, str2){
    var diff = Date.parse(str2) - Date.parse(str1); 
    return isNaN(diff) ? NaN : {
        diff: diff,
        ms: Math.ceil(diff % 1000),
        s: Math.ceil(diff / 1000 % 60),
        m: Math.ceil(diff / 60000 % 60),
        h: Math.ceil(diff / 3600000 % 24),
        d: Math.ceil(diff / 86400000)
    };
}

更好的解决方案

忽略时间部分

如果两个日期相同,则返回0。

function dayDiff(firstDate, secondDate) { firstDate = new Date(firstDate); secondDate = new Date(secondDate); if (!isNaN(firstDate) && !isNaN(secondDate)) { firstDate.setHours(0, 0, 0, 0); //ignore time part secondDate.setHours(0, 0, 0, 0); //ignore time part var dayDiff = secondDate - firstDate; dayDiff = dayDiff / 86400000; // divide by milisec in one day console.log(dayDiff); } else { console.log("Enter valid date."); } } $(document).ready(function() { $('input[type=datetime]').datepicker({ dateFormat: "mm/dd/yy", changeMonth: true, changeYear: true }); $("#button").click(function() { dayDiff($('#first').val(), $('#second').val()); }); }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" href="//code.jquery.com/ui/1.12.1/themes/base/jquery-ui.css"> <script src="//code.jquery.com/ui/1.12.1/jquery-ui.js"></script> <input type="datetime" id="first" value="12/28/2016" /> <input type="datetime" id="second" value="12/28/2017" /> <input type="button" id="button" value="Calculate">

使用Moment.js

Var future = moment('05/02/2015'); Var start = moment('04/23/2015'); Var d =未来。diff(开始,“天”);/ / 9 console.log (d); < script src = " https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.17.1/moment-with-locales.min.js " > < /脚本>