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

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。

其他回答

一行代码和小代码

const diff=(e,t)=>Math.floor((new Date(e).getTime()-new Date(t).getTime())/1000*60*60*24);

// or

const diff=(e,t)=>Math.floor((new Date(e)-new Date(t))/864e5);

// or

const diff=(a,b)=>(new Date(a)-new Date(b))/864e5|0; 

// use
diff('1/1/2001', '1/1/2000')

为打印稿

const diff = (from: string, to: string) => Math.floor((new Date(from).getTime() - new Date(to).getTime()) / 86400000);

从DatePicker小部件使用formatDate怎么样?您可以使用它来转换时间戳格式的日期(从01/01/1970开始的毫秒),然后做一个简单的减法。

   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;
        }
    }

试试这个

let today = new Date(). toisostring()。片(0,10) const startDate = '2021-04-15'; const endDate = today; const diffInMs = new Date(endDate) - new Date(startDate) const diffInDays = diffInMs / (1000 * 60 * 60 * 24); alert(diffInDays);

我使用下面的代码来试验新闻帖子的发布日期功能。我根据发布日期和当前日期计算分钟、小时、天或年。

var startDate= new Date("Mon Jan 01 2007 11:00:00");
var endDate  =new Date("Tue Jan 02 2007 12:50:00");
var timeStart = startDate.getTime();
var timeEnd = endDate.getTime();
var yearStart = startDate.getFullYear();
var yearEnd   = endDate.getFullYear();
if(yearStart == yearEnd)
 {
  var hourDiff = timeEnd - timeStart; 
  var secDiff = hourDiff / 1000;
  var minDiff = hourDiff / 60 / 1000; 
  var hDiff = hourDiff / 3600 / 1000; 
  var myObj = {};
  myObj.hours = Math.floor(hDiff);
  myObj.minutes = minDiff  
  if(myObj.hours >= 24)
   {
    console.log(Math.floor(myObj.hours/24) + "day(s) ago")
   } 
 else if(myObj.hours>0)
  {
   console.log(myObj.hours +"hour(s) ago")
  }
 else
  {
   console.log(Math.abs(myObj.minutes) +"minute(s) ago")
  }
}
else
{
var yearDiff = yearEnd - yearStart;
console.log( yearDiff +" year(s) ago");
}