119

我正在创建一个应用程序，它允许您定义具有时间范围的事件。我想在用户选择或更改开始日期时自动填写结束日期。但是，我不太清楚如何获得两次之间的差异，然后如何使用该差异创建新的结束日期。

4

16 回答 16

79

getTime()在 JavaScript 中，可以通过调用方法或仅使用数字表达式中的日期将日期转换为自 epoc 以来的毫秒数。

因此，要获得差异，只需减去两个日期。

要根据差异创建新日期，只需在构造函数中传递毫秒数。

var oldBegin = ...
var oldEnd = ...
var newBegin = ...

var newEnd = new Date(newBegin + oldEnd - oldBegin);

这应该只是工作

编辑：修复了@bdukes 指出的错误

编辑：

对于行为的解释oldBegin，oldEnd、 和newBegin是Date实例。调用操作符+并将-触发 Javascript 自动转换并自动调用valueOf()这些对象的原型方法。碰巧该valueOf()方法在Date对象中实现为对getTime().

所以基本上：date.getTime() === date.valueOf() === (0 + date) === (+date)

于 2008-09-03T15:42:01.787 回答
27

JavaScript 完美支持开箱即用的日期差异

https://jsfiddle.net/b9chris/v5twbe3h/

var msMinute = 60*1000, 
    msDay = 60*60*24*1000,
    a = new Date(2012, 2, 12, 23, 59, 59),
    b = new Date("2013 march 12");


console.log(Math.floor((b - a) / msDay) + ' full days between'); // 364
console.log(Math.floor(((b - a) % msDay) / msMinute) + ' full minutes between'); // 0

现在有些陷阱。试试这个：

console.log(a - 10); // 1331614798990
console.log(a + 10); // mixed string

因此，如果您有添加数字和日期的风险，请number直接将日期转换为。

console.log(a.getTime() - 10); // 1331614798990
console.log(a.getTime() + 10); // 1331614799010

我的第一个示例演示了 Date 对象的强大功能，但它实际上似乎是一个定时炸弹

于 2013-04-04T12:08:30.017 回答
10

见 JsFiddle 演示

    var date1 = new Date();    
    var date2 = new Date("2025/07/30 21:59:00");
    //Customise date2 for your required future time

    showDiff();

function showDiff(date1, date2){

    var diff = (date2 - date1)/1000;
    diff = Math.abs(Math.floor(diff));

    var days = Math.floor(diff/(24*60*60));
    var leftSec = diff - days * 24*60*60;

    var hrs = Math.floor(leftSec/(60*60));
    var leftSec = leftSec - hrs * 60*60;

    var min = Math.floor(leftSec/(60));
    var leftSec = leftSec - min * 60;

    document.getElementById("showTime").innerHTML = "You have " + days + " days " + hrs + " hours " + min + " minutes and " + leftSec + " seconds before death.";

setTimeout(showDiff,1000);
}

对于您的 HTML 代码：

<div id="showTime"></div>
于 2014-07-30T16:27:22.773 回答
5

如果您不关心时间组件，则可以使用.getDate()and.setDate()来设置日期部分。

因此，要将结束日期设置为开始日期后 2 周，请执行以下操作：

function GetEndDate(startDate)
{
    var endDate = new Date(startDate.getTime());
    endDate.setDate(endDate.getDate()+14);
    return endDate;
}

要返回两个日期之间的差异（以天为单位），请执行以下操作：

function GetDateDiff(startDate, endDate)
{
    return endDate.getDate() - startDate.getDate();
}

最后，让我们修改第一个函数，使其可以将 2nd 返回的值作为参数：

function GetEndDate(startDate, days)
{
    var endDate = new Date(startDate.getTime());
    endDate.setDate(endDate.getDate() + days);
    return endDate;
}
于 2008-09-03T15:51:56.570 回答
4

谢谢@Vincent Robert，我最终使用了您的基本示例，尽管它实际上是newBegin + oldEnd - oldBegin. 这是简化的最终解决方案：

    // don't update end date if there's already an end date but not an old start date
    if (!oldEnd || oldBegin) {
        var selectedDateSpan = 1800000; // 30 minutes
        if (oldEnd) {
            selectedDateSpan = oldEnd - oldBegin;
        }

       newEnd = new Date(newBegin.getTime() + selectedDateSpan));
    }
于 2008-09-03T16:38:43.930 回答
3

根据您的需要，此函数将计算 2 天之间的差异，并返回以天为小数的结果。

// This one returns a signed decimal. The sign indicates past or future.

this.getDateDiff = function(date1, date2) {
    return (date1.getTime() - date2.getTime()) / (1000 * 60 * 60 * 24);
}

// This one always returns a positive decimal. (Suggested by Koen below)

this.getDateDiff = function(date1, date2) {
    return Math.abs((date1.getTime() - date2.getTime()) / (1000 * 60 * 60 * 24));
}
于 2012-05-10T14:39:41.993 回答
2

如果使用 moment.js，有一个更简单的解决方案，它可以在一行代码中为您提供天数差异。

moment(endDate).diff(moment(beginDate), 'days');

可以在moment.js 页面中找到更多详细信息

干杯，米格尔

于 2016-05-30T19:51:02.403 回答
1 
function compare()
{
  var end_actual_time    = $('#date3').val();

  start_actual_time = new Date();
  end_actual_time = new Date(end_actual_time);

  var diff = end_actual_time-start_actual_time;

  var diffSeconds = diff/1000;
  var HH = Math.floor(diffSeconds/3600);
  var MM = Math.floor(diffSeconds%3600)/60;

  var formatted = ((HH < 10)?("0" + HH):HH) + ":" + ((MM < 10)?("0" + MM):MM)
  getTime(diffSeconds);
}
function getTime(seconds) {
  var days = Math.floor(leftover / 86400);

  //how many seconds are left
  leftover = leftover - (days * 86400);

  //how many full hours fits in the amount of leftover seconds
  var hours = Math.floor(leftover / 3600);

  //how many seconds are left
  leftover = leftover - (hours * 3600);

  //how many minutes fits in the amount of leftover seconds
  var minutes = leftover / 60;

  //how many seconds are left
  //leftover = leftover - (minutes * 60);
  alert(days + ':' + hours + ':' + minutes);
}
于 2013-11-15T13:19:37.367 回答
1

替代修改扩展代码..

http://jsfiddle.net/vvGPQ/48/

showDiff();

function showDiff(){
var date1 = new Date("2013/01/18 06:59:00");   
var date2 = new Date();
//Customise date2 for your required future time

var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));

var years = Math.floor(diff/(365*24*60*60));
var leftSec = diff - years * 365*24*60*60;

var month = Math.floor(leftSec/((365/12)*24*60*60));
var leftSec = leftSec - month * (365/12)*24*60*60;    

var days = Math.floor(leftSec/(24*60*60));
var leftSec = leftSec - days * 24*60*60;

var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;

var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;




document.getElementById("showTime").innerHTML = "You have " + years + " years "+ month + " month " + days + " days " + hrs + " hours " + min + " minutes and " + leftSec + " seconds the life time has passed.";

setTimeout(showDiff,1000);
}
于 2015-02-19T20:22:15.987 回答
1 
<html>
<head>
<script>
function dayDiff()
{
     var start = document.getElementById("datepicker").value;
     var end= document.getElementById("date_picker").value;
     var oneDay = 24*60*60*1000; 
     var firstDate = new Date(start);
     var secondDate = new Date(end);    
     var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
    document.getElementById("leave").value =diffDays ;
 }
</script>
</head>
<body>
<input type="text" name="datepicker"value=""/>
<input type="text" name="date_picker" onclick="function dayDiff()" value=""/>
<input type="text" name="leave" value=""/>
</body>
</html>
于 2016-01-14T06:04:32.953 回答
1

下面的代码将返回从今天到期货日期的剩余天数。

依赖项：jQuery 和 MomentJs。

var getDaysLeft = function (date) {
  var today = new Date();
  var daysLeftInMilliSec = Math.abs(new Date(moment(today).format('YYYY-MM-DD')) - new Date(date));
  var daysLeft = daysLeftInMilliSec / (1000 * 60 * 60 * 24);   
  return daysLeft;
};

getDaysLeft('YYYY-MM-DD');
于 2018-03-30T14:15:53.010 回答
1 
var getDaysLeft = function (date1, date2) {
   var daysDiffInMilliSec = Math.abs(new Date(date1) - new Date(date2));
   var daysLeft = daysDiffInMilliSec / (1000 * 60 * 60 * 24);   
   return daysLeft;
};
var date1='2018-05-18';
var date2='2018-05-25';
var dateDiff = getDaysLeft(date1, date2);
console.log(dateDiff);
于 2018-05-18T15:35:35.260 回答
0

如果您使用 Date 对象，然后getTime()对两个日期都使用该函数，它将以数值形式为您提供自 1970 年 1 月 1 日以来各自的时间。然后，您可以得到这些数字之间的差异。

如果这对您没有帮助，请查看完整的文档：http ://www.w3schools.com/jsref/jsref_obj_date.asp

于 2008-09-03T15:46:14.290 回答
0 
function checkdate() {
    var indate = new Date()
    indate.setDate(dat)
    indate.setMonth(mon - 1)
    indate.setFullYear(year)

    var one_day = 1000 * 60 * 60 * 24
    var diff = Math.ceil((indate.getTime() - now.getTime()) / (one_day))
    var str = diff + " days are remaining.."
    document.getElementById('print').innerHTML = str.fontcolor('blue')
}
于 2011-02-20T11:23:17.243 回答
0

当您输入学习的开始日期和结束日期（合格的准确日期）并检查持续时间是否小于一年时，此代码将填写学习年限，如果是，则提醒消息注意有三个输入元素，第一个txtFromQualifDate和第二txtQualifDate和第三txtStudyYears

它将显示小数年数的结果

function getStudyYears()
    {
        if(document.getElementById('txtFromQualifDate').value != '' && document.getElementById('txtQualifDate').value != '')
        {
            var d1 = document.getElementById('txtFromQualifDate').value;

            var d2 = document.getElementById('txtQualifDate').value;

            var one_day=1000*60*60*24;

            var x = d1.split("/");
            var y = d2.split("/");

            var date1=new Date(x[2],(x[1]-1),x[0]);

            var date2=new Date(y[2],(y[1]-1),y[0])

            var dDays = (date2.getTime()-date1.getTime())/one_day;

            if(dDays < 365)
            {
                alert("the date between start study and graduate must not be less than a year !");

                document.getElementById('txtQualifDate').value = "";
                document.getElementById('txtStudyYears').value = "";

                return ;
            }

            var dMonths = Math.ceil(dDays / 30);

            var dYears = Math.floor(dMonths /12) + "." + dMonths % 12;

            document.getElementById('txtStudyYears').value = dYears;
        }
    }
于 2012-05-02T07:03:12.103 回答
0

这就是我在我的系统上所做的。

var startTime=("08:00:00").split(":");
var endTime=("16:00:00").split(":");
var HoursInMinutes=((parseInt(endTime[0])*60)+parseInt(endTime[1]))-((parseInt(startTime[0])*60)+parseInt(startTime[1]));
console.log(HoursInMinutes/60);
于 2018-05-24T11:32:53.823 回答