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。

下面是datediff的快速实现，作为解决问题的概念证明。它依赖于这样一个事实，即您可以通过减去两个日期之间经过的毫秒，这将它们强制转换为原始数字值(自1970年初以来的毫秒)。

/** * Take the difference between the dates and divide by milliseconds per day. * Round to nearest whole number to deal with DST. */ function datediff(first, second) { return Math.round((second - first) / (1000 * 60 * 60 * 24)); } /** * new Date("dateString") is browser-dependent and discouraged, so we'll write * a simple parse function for U.S. date format (which does no error checking) */ function parseDate(str) { var mdy = str.split('/'); return new Date(mdy[2], mdy[0] - 1, mdy[1]); } alert(datediff(parseDate(first.value), parseDate(second.value))); <input id="first" value="1/1/2000"/> <input id="second" value="1/1/2001"/>

You should be aware that the "normal" Date APIs (without "UTC" in the name) operate in the local timezone of the user's browser, so in general you could run into issues if your user is in a timezone that you don't expect, and your code will have to deal with Daylight Saving Time transitions. You should carefully read the documentation for the Date object and its methods, and for anything more complicated, strongly consider using a library that offers more safe and powerful APIs for date manipulation.

数字和日期——MDN JavaScript指南 日期——MDN JavaScript参考

同样，出于说明的目的，为了简洁起见，该代码段对窗口对象使用了命名访问，但在生产中应该使用getElementById之类的标准化api，或者更有可能使用一些UI框架。

我来找这个小工具在里面你会找到这个的函数。这里有一个简短的例子:

        <script type="text/javascript" src="date.js"></script>
        <script type="text/javascript">
            var minutes = 1000*60;
            var hours = minutes*60;
            var days = hours*24;

            var foo_date1 = getDateFromFormat("02/10/2009", "M/d/y");
            var foo_date2 = getDateFromFormat("02/12/2009", "M/d/y");

            var diff_date = Math.round((foo_date2 - foo_date1)/days);
            alert("Diff date is: " + diff_date );
        </script>

区分两个日期的最简单方法是:

var diff = Math.floor((Date.parse(str2) - Date.parse(str1)) / 86400000);

您将得到不同的天数(如果其中一个或两个都无法解析，则为NaN)。解析日期给出了以毫秒为单位的结果，要按天得到它，你必须除以24 * 60 * 60 * 1000

如果你想用天、小时、分钟、秒和毫秒来划分:

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

以下是我对James版本的重构版本:

function mydiff(date1,date2,interval) {
    var second=1000, minute=second*60, hour=minute*60, day=hour*24, week=day*7;
    date1 = new Date(date1);
    date2 = new Date(date2);
    var timediff = date2 - date1;
    if (isNaN(timediff)) return NaN;
    switch (interval) {
        case "years": return date2.getFullYear() - date1.getFullYear();
        case "months": return (
            ( date2.getFullYear() * 12 + date2.getMonth() )
            -
            ( date1.getFullYear() * 12 + date1.getMonth() )
        );
        case "weeks"  : return Math.floor(timediff / week);
        case "days"   : return Math.floor(timediff / day); 
        case "hours"  : return Math.floor(timediff / hour); 
        case "minutes": return Math.floor(timediff / minute);
        case "seconds": return Math.floor(timediff / second);
        default: return undefined;
    }
}

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)
    };
}

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。

当我想在两个日期上做一些计算时，我发现了这个问题，但是日期有小时和分钟的值，我修改了@michael-liu的答案来满足我的要求，它通过了我的测试。

差异日期2012-12-31 23:00和2013-01-01 01:00应该等于1。(2小时) 差异日期2012-12-31 01:00和2013-01-01 23:00应该等于1。(46个小时)

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

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

我建议使用moment.js库(http://momentjs.com/docs/#/displaying/difference/)。它正确地处理夏令时，通常是很好的工作。

例子:

var start = moment("2013-11-03");
var end = moment("2013-11-04");
end.diff(start, "days")
1

JS中的日期值是datetime值。

因此，直接日期计算是不一致的:

(2013-11-05 00:00:00) - (2013-11-04 10:10:10) < 1 day

例如，我们需要转换第2个日期:

(2013-11-05 00:00:00) - (2013-11-04 00:00:00) = 1 day

该方法可以在两个日期截断轧机:

var date1 = new Date('2013/11/04 00:00:00'); var date2 = new Date('2013/11/04 10:10:10'); //less than 1 var start = Math.floor(date1.getTime() / (3600 * 24 * 1000)); //days as integer from.. var end = Math.floor(date2.getTime() / (3600 * 24 * 1000)); //days as integer from.. var daysDiff = end - start; // exact dates console.log(daysDiff); date2 = new Date('2013/11/05 00:00:00'); //1 var start = Math.floor(date1.getTime() / (3600 * 24 * 1000)); //days as integer from.. var end = Math.floor(date2.getTime() / (3600 * 24 * 1000)); //days as integer from.. var daysDiff = end - start; // exact dates console.log(daysDiff);

Date.prototype.days =函数(到){ 返回Math.abs(Math.floor(to.getTime() / (3600 * 24 * 1000)) - Math.floor(this.getTime() / (3600 * 24 * 1000))) ｝ console.log(新日期(“2014/05/20”)。天(新日期(2014/05/23)));// 3天 console.log(新日期(“2014/05/23”)。天(新日期(2014/05/20)));// 3天

这可能不是最优雅的解决方案，但我认为它似乎用一段相对简单的代码就回答了这个问题。你不能用这样的词吗?

function dayDiff(startdate, enddate) {
  var dayCount = 0;

  while(enddate >= startdate) {
    dayCount++;
    startdate.setDate(startdate.getDate() + 1);
  }

return dayCount; 
}

这是假设您将日期对象作为参数传递。

最好还是取消夏令时吧，马斯。装天花板,数学。楼层等，使用UTC时间:

var firstDate = Date.UTC(2015,01,2);
var secondDate = Date.UTC(2015,04,22);
var diff = Math.abs((firstDate.valueOf() 
    - secondDate.valueOf())/(24*60*60*1000));

这个例子给出了109天的差异。24*60*60*1000是一天，单位是毫秒。

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>

我在Angular中也遇到了同样的问题。我复制了一份，否则他会覆盖第一次约会。两个日期的时间都必须为00:00:00(显然)

 /*
* Deze functie gebruiken we om het aantal dagen te bereken van een booking.
* */
$scope.berekenDagen = function ()
{
    $scope.booking.aantalDagen=0;

    /*De loper is gelijk aan de startdag van je reservatie.
     * De copy is nodig anders overschijft angular de booking.van.
     * */
    var loper = angular.copy($scope.booking.van);

    /*Zolang de reservatie beschikbaar is, doorloop de weekdagen van je start tot einddatum.*/
    while (loper < $scope.booking.tot) {
        /*Tel een dag op bij je loper.*/
        loper.setDate(loper.getDate() + 1);
        $scope.booking.aantalDagen++;
    }

    /*Start datum telt natuurlijk ook mee*/
    $scope.booking.aantalDagen++;
    $scope.infomsg +=" aantal dagen: "+$scope.booking.aantalDagen;
};

   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;
        }
    }

如果你有两个unix时间戳，你可以使用这个函数(为了清晰起见，写得更详细一点):

// Calculate number of days between two unix timestamps
// ------------------------------------------------------------
var daysBetween = function(timeStampA, timeStampB) {
    var oneDay = 24 * 60 * 60 * 1000; // hours * minutes * seconds * milliseconds
    var firstDate = new Date(timeStampA * 1000);
    var secondDate = new Date(timeStampB * 1000);
    var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
    return diffDays;
};

例子:

daysBetween(1096580303, 1308713220); // 2455

要计算两个给定日期之间的天数，可以使用以下代码。我在这里使用的日期是2016年1月1日和2016年12月31日

var day_start = new日期(“2016年1月1日”); var day_end =新的日期(“2016年12月31日”); Var total_days = (day_end - day_start) / (1000 * 60 * 60 * 24); . getelementbyid(“演示”)。innerHTML = Math.round(total_days); <h3>天之间的给定日期</h3> < p id = "演示" > < / p >

使用以下公式可以计算出两个不同TZs的休止日期之间的完整证明天数差:

var start = new Date('10/3/2015'); var end = new Date('11/2/2015'); Var天数= (end - start) / 1000 / 60 / 60 / 24; console.log(天); //实际上是30;但由于夏令时，将显示31.0xxx //你需要抵消如下 days = days - (end.getTimezoneOffset() - start.getTimezoneOffset()) / (60 * 24); console.log(天);

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");
}

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

使用毫秒时要小心。

date.getTime()返回毫秒，用毫秒做数学运算需要包含

日光节约时间(DST) 检查两个日期的时间是否相同(小时，分钟，秒，毫秒) 请确定需要哪些天数差异:2016年9月19日- 2016年9月29日= 1天或2天的差异?

上面评论中的例子是我迄今为止找到的最好的解决方案 https://stackoverflow.com/a/11252167/2091095。但是，如果你想计算所有涉及的天数，则对其结果使用+1。

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;
}

var diff = daysBetween($('#first').val(), $('#second').val()) + 1;

var start= $("#firstDate").datepicker("getDate");
var end= $("#SecondDate").datepicker("getDate");
var days = (end- start) / (1000 * 60 * 60 * 24);
 alert(Math.round(days));

Jsfiddle示例:)

function timeDifference(date1, date2) { var oneDay = 24 * 60 * 60; // hours*minutes*seconds var oneHour = 60 * 60; // minutes*seconds var oneMinute = 60; // 60 seconds var firstDate = date1.getTime(); // convert to milliseconds var secondDate = date2.getTime(); // convert to milliseconds var seconds = Math.round(Math.abs(firstDate - secondDate) / 1000); //calculate the diffrence in seconds // the difference object var difference = { "days": 0, "hours": 0, "minutes": 0, "seconds": 0, } //calculate all the days and substract it from the total while (seconds >= oneDay) { difference.days++; seconds -= oneDay; } //calculate all the remaining hours then substract it from the total while (seconds >= oneHour) { difference.hours++; seconds -= oneHour; } //calculate all the remaining minutes then substract it from the total while (seconds >= oneMinute) { difference.minutes++; seconds -= oneMinute; } //the remaining seconds : difference.seconds = seconds; //return the difference object return difference; } console.log(timeDifference(new Date(2017,0,1,0,0,0),new Date()));

function formatDate(seconds, dictionary) {
    var foo = new Date;
    var unixtime_ms = foo.getTime();
    var unixtime = parseInt(unixtime_ms / 1000);
    var diff = unixtime - seconds;
    var display_date;
    if (diff <= 0) {
        display_date = dictionary.now;
    } else if (diff < 60) {
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.second;
        } else {
            display_date = diff + ' ' + dictionary.seconds;
        }
    } else if (diff < 3540) {
        diff = Math.round(diff / 60);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.minute;
        } else {
            display_date = diff + ' ' + dictionary.minutes;
        }
    } else if (diff < 82800) {
        diff = Math.round(diff / 3600);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.hour;
        } else {
            display_date = diff + ' ' + dictionary.hours;
        }
    } else {
        diff = Math.round(diff / 86400);
        if (diff == 1) {
            display_date = diff + ' ' + dictionary.day;
        } else {
            display_date = diff + ' ' + dictionary.days;
        }
    }
    return display_date;
}

您可以使用UnderscoreJS来格式化和计算差异。

演示https://jsfiddle.net/sumitridhal/8sv94msp/

var startDate = moment(“206 -08- 29t23:35:01”); var endDate = moment(“206 -08- 30t23:35:01”); 游戏机。log (startDate); 游戏机。log (endDate); var结果= endDate。diff(startDate, hours, true); 文件。 文档全身appendChild(文档。createTextNode (resultHours)); 身体(白色空间:pre;font-family: monospace;) “https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.5.1/moment.min.js”< script src = > / < script >

Bookmarklet版本的其他答案，提示你两个日期:

javascript:(function() {
    var d = new Date(prompt("First Date or leave blank for today?") || Date.now());
    prompt("Days Between", Math.round(
        Math.abs(
            (d.getTime() - new Date(prompt("Date 2")).getTime())
                /(24*60*60*1000)
             )
        ));
})();

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);

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">

这个答案基于另一个答案(链接在最后)，是关于两个日期之间的差异。 你可以看到它是如何工作的，因为它很简单，它还包括将差异分成 时间单位(我做的一个函数)并转换为UTC以停止时区问题。

function date_units_diff(a, b, unit_amounts) { var split_to_whole_units = function (milliseconds, unit_amounts) { // unit_amounts = list/array of amounts of milliseconds in a // second, seconds in a minute, etc., for example "[1000, 60]". time_data = [milliseconds]; for (i = 0; i < unit_amounts.length; i++) { time_data.push(parseInt(time_data[i] / unit_amounts[i])); time_data[i] = time_data[i] % unit_amounts[i]; }; return time_data.reverse(); }; if (unit_amounts == undefined) { unit_amounts = [1000, 60, 60, 24]; }; var utc_a = new Date(a.toUTCString()); var utc_b = new Date(b.toUTCString()); var diff = (utc_b - utc_a); return split_to_whole_units(diff, unit_amounts); } // Example of use: var d = date_units_diff(new Date(2010, 0, 1, 0, 0, 0, 0), new Date()).slice(0,-2); document.write("In difference: 0 days, 1 hours, 2 minutes.".replace( /0|1|2/g, function (x) {return String( d[Number(x)] );} ));

我上面的代码是如何工作的

日期/时间差异，以毫秒为单位，可以使用date对象计算:

var a = new Date(); // Current date now.
var b = new Date(2010, 0, 1, 0, 0, 0, 0); // Start of 2010.

var utc_a = new Date(a.toUTCString());
var utc_b = new Date(b.toUTCString());
var diff = (utc_b - utc_a); // The difference as milliseconds.

然后算出这个差值的秒数，将其除以1000进行换算 毫秒到秒，然后将结果更改为整数(整数)以删除 毫秒数(小数的小数部分):var seconds = parseInt(diff/1000)。 此外，我可以使用相同的过程获得更长的时间单位，例如: -(整)分钟，秒除以60，结果变为整数， —hours，分钟除以60，返回结果为整数。

我创建了一个函数来完成这个过程，把差值分成 整个时间单位，命名为split_to_whole_units，演示如下:

console.log(split_to_whole_units(72000, [1000, 60]));
// -> [1,12,0] # 1 (whole) minute, 12 seconds, 0 milliseconds.

这个答案是基于另一个答案的。

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

我只有两个以毫秒为单位的时间戳，所以我必须用moment.js做一些额外的步骤来获得天数。

const getDaysDiff = (fromTimestamp, toTimestamp) => {
    // set timezone offset with utcOffset if needed
    let fromDate = moment(fromTimestamp).utcOffset(8);
    let toDate = moment(toTimestamp).utcOffset(8);
    // get the start moment of the day
    fromDate.set({'hour':0, 'minute': 0, 'second': 0, 'millisecond': 0});
    toDate.set({'hour':0, 'minute': 0, 'second': 0, 'millisecond': 0});
    let diffDays = toDate.diff(fromDate, 'days');

    return diffDays;
}

getDaysDiff(1528889400000, 1528944180000)// 1 

我也有同样的问题，但如果你在SQL查询上做的话会更好:

DateDiff(DAY, StartValue,GETDATE()) AS CountDays

查询将自动生成一个列CountDays

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);
  })();
}

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>

简单、容易、复杂。此函数将每1秒调用一次以更新时间。

const year = (new Date().getFullYear());
const bdayDate = new Date("04,11,2019").getTime(); //mmddyyyy

// countdown
let timer = setInterval(function () {

    // get today's date
    const today = new Date().getTime();

    // get the difference
    const diff = bdayDate - today;

    // math
    let days = Math.floor(diff / (1000 * 60 * 60 * 24));
    let hours = Math.floor((diff % (1000 * 60 * 60 * 24)) / (1000 * 60 * 60));
    let minutes = Math.floor((diff % (1000 * 60 * 60)) / (1000 * 60));
    let seconds = Math.floor((diff % (1000 * 60)) / 1000);

}, 1000);

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

如果我们想计算我们的年龄，这是一个有点不同的答案

    {
      birthday: 'April 22, 1993',
      names: {
        first: 'Keith',
        last: 'Buckley'
      }
    },
    {
      birthday: 'January 3, 1975',
      names: {
        first: 'Larry',
        last: 'Heep'
      }
    },
    {
      birthday: 'February 12, 1944',
      names: {
        first: 'Linda',
        last: 'Bermeer'
      }
    }
  ];
const cleanPeople = people.map(function ({birthday, names:{first, last}}) {
      // birthday, age, fullName;
      const now = new Date();
      var age =  Math.floor(( Date.parse(now) - Date.parse(birthday)) / 31536000000);
      return {
        age,
        fullName:`${first} ${last}`
      }
    });
    console.log(cleanPeople);
    console.table(cleanPeople);

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);

我最近也有同样的问题，因为我来自Java世界，所以我立即开始搜索JavaScript的JSR 310实现。JSR 310是Java的日期和时间API (Java 8的标准版本)。我认为这个API设计得很好。

幸运的是，有一个直接的Javascript端口，称为js-joda。

首先，在<head>中包含js-joda:

<script
    src="https://cdnjs.cloudflare.com/ajax/libs/js-joda/1.11.0/js-joda.min.js"
    integrity="sha512-piLlO+P2f15QHjUv0DEXBd4HvkL03Orhi30Ur5n1E4Gk2LE4BxiBAP/AD+dxhxpW66DiMY2wZqQWHAuS53RFDg=="
    crossorigin="anonymous"></script>

然后简单地这样做:

let date1 = JSJoda.LocalDate.of(2020, 12, 1);
let date2 = JSJoda.LocalDate.of(2021, 1, 1);
let daysBetween = JSJoda.ChronoUnit.DAYS.between(date1, date2);

现在daysBetween包含了间隔的天数。注意，结束日期是独占的。

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);

 // JavaScript / NodeJs answer  
   let startDate = new Date("2022-09-19");
   let endDate = new Date("2022-09-26");

   let difference = startDate.getTime() - endDate.getTime();
   
    console.log(difference);

   let TotalDiffDays = Math.ceil(difference / (1000 * 3600 * 24));
   console.log(TotalDiffDays + " days :) ");

夏令时问题使这里的许多答案无效。我将使用一个helper函数来获得给定日期的唯一天数——通过使用UTC方法:

const dayNumber = a => Date.UTC(a.getFullYear(), a.getMonth(), a.getDate()) / (24*60*60*1000); const daysBetween = (a, b) => dayNumber(b) - dayNumber(a); // Testing it const start = new Date(1000, 0, 1); // 1 January 1000 const end = new Date(3000, 0, 1); // 1 January 3000 let current = new Date(start); for (let days = 0; current < end; days++) { const diff = daysBetween(start, current); if (diff !== days) throw "test failed"; current.setDate(current.getDate() + 1); // move current date one day forward } console.log("tests succeeded");