JavaScript中几年,几个月,几天的两个日期之间的差异

Chr*_*ris 41 javascript datediff date

我现在已经搜索了4个小时,并且没有找到解决方案来获取JavaScript中年,月和日的两个日期之间的差异,例如:2010年4月10日是3年,x个月和y天前.

有很多解决方案,但它们只提供日期或月份或年份的格式差异,或者它们不正确(意味着不处理一个月或闰年的实际天数等).这样做真的很难吗?

我看过:

在PHP中它很容易,但不幸的是我只能在该项目上使用客户端脚本.任何能够做到这一点的库或框架都可以.

以下是日期差异的预期输出列表:

//Expected output should be: "1 year, 5 months".
diffDate(new Date('2014-05-10'), new Date('2015-10-10'));

//Expected output should be: "1 year, 4 months, 29 days".
diffDate(new Date('2014-05-10'), new Date('2015-10-09'));

//Expected output should be: "1 year, 3 months, 30 days".
diffDate(new Date('2014-05-10'), new Date('2015-09-09'));

//Expected output should be: "9 months, 27 days".
diffDate(new Date('2014-05-10'), new Date('2015-03-09'));

//Expected output should be: "1 year, 9 months, 28 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-09'));

//Expected output should be: "1 year, 10 months, 1 days".
diffDate(new Date('2014-05-10'), new Date('2016-03-11'));
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Paw*_*ech 29

你需要多精确?如果你确实需要考虑常见的年份和闰年,以及几个月之间天数的确切差异,那么你将不得不写一些更高级的东西,但是为了进行基本和粗略的计算,这应该可以解决问题:

today = new Date()
past = new Date(2010,05,01) // remember this is equivalent to 06 01 2010
//dates in js are counted from 0, so 05 is june

function calcDate(date1,date2) {
    var diff = Math.floor(date1.getTime() - date2.getTime());
    var day = 1000 * 60 * 60 * 24;

    var days = Math.floor(diff/day);
    var months = Math.floor(days/31);
    var years = Math.floor(months/12);

    var message = date2.toDateString();
    message += " was "
    message += days + " days " 
    message += months + " months "
    message += years + " years ago \n"

    return message
    }


a = calcDate(today,past)
console.log(a) // returns Tue Jun 01 2010 was 1143 days 36 months 3 years ago
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请记住,这是不精确的,为了计算完全精确的日期,必须有一个日历并知道一年是否是闰年,我计算月数的方式也只是近似值.

但你可以轻松改进它.

  • 感谢您的回答.但我修改了一点,下面给出了这个链接:http://jsfiddle.net/PUSQU/8 (2认同)
  • 这是一个很好的答案,但令我惊讶的是,没有人考虑到这有些不对劲……因为月份和年份的数学计算将基于有 372 天的一年 (2认同)

Ina*_*mus 18

实际上,有一个带有moment.js插件的解决方案,这很简单.

你可以使用moment.js

不要再重新发明轮子.

只需插入Moment.js日期范围插件即可.


例:

var starts = moment('2014-02-03 12:53:12');
var ends   = moment();

var duration = moment.duration(ends.diff(starts));

// with ###moment precise date range plugin###
// it will tell you the difference in human terms

var diff = moment.preciseDiff(starts, ends, true); 
// example: { "years": 2, "months": 7, "days": 0, "hours": 6, "minutes": 29, "seconds": 17, "firstDateWasLater":  false }


// or as string:
var diffHuman = moment.preciseDiff(starts, ends);
// example: 2 years 7 months 6 hours 29 minutes 17 seconds

document.getElementById('output1').innerHTML = JSON.stringify(diff)
document.getElementById('output2').innerHTML = diffHuman
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<html>
<head>

  <script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.14.1/moment.min.js"></script>

  <script src="https://raw.githubusercontent.com/codebox/moment-precise-range/master/moment-precise-range.js"></script>

</head>
<body>
  
  <h2>Difference between "NOW and 2014-02-03 12:53:12"</h2>
  <span id="output1"></span>
  <br />
  <span id="output2"></span>
  
</body>
</html>
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  • 如果您需要的是2线功能,您需要支付超过100k的罚款 (3认同)
  • 压缩版只有51KB。如果服务器托管启用了gzip压缩(〜cloudflare托管),则约为20KB。更不用说浏览器可能已经从您访问过的另一个站点缓存了,该站点正在使用相同的CDN参考。考虑到该库还可以对日期进行其他许多有用的处理,考虑到JavaScript中处理日期的难度,我认为这是一个很好的折衷方案。 (3认同)
  • 那么不要将它用于这些情况。 (2认同)

Mor*_*red 16

对此进行了修改,使其更加准确。它将日期转换为 'YYYY-MM-DD' 格式,忽略 HH:MM:SS,并采用可选的 endDate 或使用当前日期,并且不关心值的顺序。

function dateDiff(startingDate, endingDate) {
    var startDate = new Date(new Date(startingDate).toISOString().substr(0, 10));
    if (!endingDate) {
        endingDate = new Date().toISOString().substr(0, 10);    // need date in YYYY-MM-DD format
    }
    var endDate = new Date(endingDate);
    if (startDate > endDate) {
        var swap = startDate;
        startDate = endDate;
        endDate = swap;
    }
    var startYear = startDate.getFullYear();
    var february = (startYear % 4 === 0 && startYear % 100 !== 0) || startYear % 400 === 0 ? 29 : 28;
    var daysInMonth = [31, february, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];

    var yearDiff = endDate.getFullYear() - startYear;
    var monthDiff = endDate.getMonth() - startDate.getMonth();
    if (monthDiff < 0) {
        yearDiff--;
        monthDiff += 12;
    }
    var dayDiff = endDate.getDate() - startDate.getDate();
    if (dayDiff < 0) {
        if (monthDiff > 0) {
            monthDiff--;
        } else {
            yearDiff--;
            monthDiff = 11;
        }
        dayDiff += daysInMonth[startDate.getMonth()];
    }

    return yearDiff + 'Y ' + monthDiff + 'M ' + dayDiff + 'D';
}
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然后你可以像这样使用它:

// based on a current date of 2019-05-10
dateDiff('2019-05-10'); // 0Y 0M 0D
dateDiff('2019-05-09'); // 0Y 0M 1D
dateDiff('2018-05-09'); // 1Y 0M 1D
dateDiff('2018-05-18'); // 0Y 11M 23D
dateDiff('2019-01-09'); // 0Y 4M 1D
dateDiff('2019-02-10'); // 0Y 3M 0D
dateDiff('2019-02-11'); // 0Y 2M 27D
dateDiff('2016-02-11'); // 3Y 2M 28D - leap year
dateDiff('1972-11-30'); // 46Y 5M 10D
dateDiff('2016-02-11', '2017-02-11'); // 1Y 0M 0D
dateDiff('2016-02-11', '2016-03-10'); // 0Y 0M 28D - leap year
dateDiff('2100-02-11', '2100-03-10'); // 0Y 0M 27D - not a leap year
dateDiff('2017-02-11', '2016-02-11'); // 1Y 0M 0D - swapped dates to return correct result
dateDiff(new Date() - 1000 * 60 * 60 * 24); // 0Y 0M 1D
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较旧的不太准确但更简单的版本

@RajeevPNadig 的答案正是我要找的,但他的代码返回的值不正确。这不是很准确,因为它假设从 1970 年 1 月 1 日开始的日期序列与具有相同天数的任何其他序列相同。例如,它将 7 月 1 日到 9 月 1 日(62 天)的差异计算为 0Y 2M 3D 而不是 0Y 2M 0D,因为 1970 年 1 月 1 日加上 62 天是 3 月 3 日。

// startDate must be a date string
function dateAgo(date) {
    var startDate = new Date(date);
    var diffDate = new Date(new Date() - startDate);
    return ((diffDate.toISOString().slice(0, 4) - 1970) + "Y " +
        diffDate.getMonth() + "M " + (diffDate.getDate()-1) + "D");
}
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然后你可以像这样使用它:

// based on a current date of 2018-03-09
dateAgo('1972-11-30'); // "45Y 3M 9D"
dateAgo('2017-03-09'); // "1Y 0M 0D"
dateAgo('2018-01-09'); // "0Y 2M 0D"
dateAgo('2018-02-09'); // "0Y 0M 28D" -- a little odd, but not wrong
dateAgo('2018-02-01'); // "0Y 1M 5D" -- definitely "feels" wrong
dateAgo('2018-03-09'); // "0Y 0M 0D"
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如果您的用例只是日期字符串,那么如果您只想要一个快速而肮脏的 4 班轮,这可以正常工作。

  • 这需要更多的赞成票。它解决了上面答案所遇到的问题。做得好! (2认同)

Raj*_*dig 12

我用这个简单的代码来区分当前日期的年,月,日.

var sdt = new Date('1972-11-30');
var difdt = new Date(new Date() - sdt);
alert((difdt.toISOString().slice(0, 4) - 1970) + "Y " + (difdt.getMonth()+1) + "M " + difdt.getDate() + "D");
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  • 编写的这段代码返回不正确的结果。假设今天的日期是 2018-03-09: `sdt = new Date("2017-03-09")` 将提醒 "1Y 1M 1D",`sdt = new Date("2018-02-09")` 将提醒"0Y 1M 29D",`sdt = new Date("2018-03-09")`将提示“0Y 1M 1D”。我对此答案的编辑被拒绝,因此我添加了更正的代码作为单独的答案。 (2认同)

Pat*_*nik 8

对于dayjs,我们是这样做的:

export const getAgeDetails = (oldDate: dayjs.Dayjs, newDate: dayjs.Dayjs) => {
  const years = newDate.diff(oldDate, 'year');
  const months = newDate.diff(oldDate, 'month') - years * 12;
  const days = newDate.diff(oldDate.add(years, 'year').add(months, 'month'), 'day');

  return {
    years,
    months,
    days,
    allDays: newDate.diff(oldDate, 'day'),
  };
};
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它完美地计算它,包括闰年和不同月份的天数。


小智 7

如果您正在使用date-fns并且不想安装 Moment.jsmoment-precise-range-plugin. 您可以使用以下date-fns函数获得与 moment-precise-range-plugin 相同的结果

intervalToDuration({
  start: new Date(),
  end: new Date("24 Jun 2020")
})
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这将在 JSON 对象中提供输出,如下所示

{
  "years": 0,
  "months": 0,
  "days": 0,
  "hours": 19,
  "minutes": 35,
  "seconds": 24
}
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实例https://stackblitz.com/edit/react-wvxvql

链接到文档https://date-fns.org/v2.14.0/docs/intervalToDuration


小智 6

为此,我创建了另一个函数:

function dateDiff(date) {
    date = date.split('-');
    var today = new Date();
    var year = today.getFullYear();
    var month = today.getMonth() + 1;
    var day = today.getDate();
    var yy = parseInt(date[0]);
    var mm = parseInt(date[1]);
    var dd = parseInt(date[2]);
    var years, months, days;
    // months
    months = month - mm;
    if (day < dd) {
        months = months - 1;
    }
    // years
    years = year - yy;
    if (month * 100 + day < mm * 100 + dd) {
        years = years - 1;
        months = months + 12;
    }
    // days
    days = Math.floor((today.getTime() - (new Date(yy + years, mm + months - 1, dd)).getTime()) / (24 * 60 * 60 * 1000));
    //
    return {years: years, months: months, days: days};
}
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不需要任何 3rd 方库。采用一个参数——YYYY-MM-DD 格式的日期。

https://gist.github.com/lemmon/d27c2d4a783b1cf72d1d1cc243458d56


Chr*_*ris 5

为了方便快捷,我前段时间写了这个函数.它以漂亮的格式返回两个日期之间的差异.随意使用它(在webkit上测试).

/**
 * Function to print date diffs.
 * 
 * @param {Date} fromDate: The valid start date
 * @param {Date} toDate: The end date. Can be null (if so the function uses "now").
 * @param {Number} levels: The number of details you want to get out (1="in 2 Months",2="in 2 Months, 20 Days",...)
 * @param {Boolean} prefix: adds "in" or "ago" to the return string
 * @return {String} Diffrence between the two dates.
 */
function getNiceTime(fromDate, toDate, levels, prefix){
    var lang = {
            "date.past": "{0} ago",
            "date.future": "in {0}",
            "date.now": "now",
            "date.year": "{0} year",
            "date.years": "{0} years",
            "date.years.prefixed": "{0} years",
            "date.month": "{0} month",
            "date.months": "{0} months",
            "date.months.prefixed": "{0} months",
            "date.day": "{0} day",
            "date.days": "{0} days",
            "date.days.prefixed": "{0} days",
            "date.hour": "{0} hour",
            "date.hours": "{0} hours",
            "date.hours.prefixed": "{0} hours",
            "date.minute": "{0} minute",
            "date.minutes": "{0} minutes",
            "date.minutes.prefixed": "{0} minutes",
            "date.second": "{0} second",
            "date.seconds": "{0} seconds",
            "date.seconds.prefixed": "{0} seconds",
        },
        langFn = function(id,params){
            var returnValue = lang[id] || "";
            if(params){
                for(var i=0;i<params.length;i++){
                    returnValue = returnValue.replace("{"+i+"}",params[i]);
                }
            }
            return returnValue;
        },
        toDate = toDate ? toDate : new Date(),
        diff = fromDate - toDate,
        past = diff < 0 ? true : false,
        diff = diff < 0 ? diff * -1 : diff,
        date = new Date(new Date(1970,0,1,0).getTime()+diff),
        returnString = '',
        count = 0,
        years = (date.getFullYear() - 1970);
    if(years > 0){
        var langSingle = "date.year" + (prefix ? "" : ""),
            langMultiple = "date.years" + (prefix ? ".prefixed" : "");
        returnString += (count > 0 ?  ', ' : '') + (years > 1 ? langFn(langMultiple,[years]) : langFn(langSingle,[years]));
        count ++;
    }
    var months = date.getMonth();
    if(count < levels && months > 0){
        var langSingle = "date.month" + (prefix ? "" : ""),
            langMultiple = "date.months" + (prefix ? ".prefixed" : "");
        returnString += (count > 0 ?  ', ' : '') + (months > 1 ? langFn(langMultiple,[months]) : langFn(langSingle,[months]));
        count ++;
    } else {
        if(count > 0)
            count = 99;
    }
    var days = date.getDate() - 1;
    if(count < levels && days > 0){
        var langSingle = "date.day" + (prefix ? "" : ""),
            langMultiple = "date.days" + (prefix ? ".prefixed" : "");
        returnString += (count > 0 ?  ', ' : '') + (days > 1 ? langFn(langMultiple,[days]) : langFn(langSingle,[days]));
        count ++;
    } else {
        if(count > 0)
            count = 99;
    }
    var hours = date.getHours();
    if(count < levels && hours > 0){
        var langSingle = "date.hour" + (prefix ? "" : ""),
            langMultiple = "date.hours" + (prefix ? ".prefixed" : "");
        returnString += (count > 0 ?  ', ' : '') + (hours > 1 ? langFn(langMultiple,[hours]) : langFn(langSingle,[hours]));
        count ++;
    } else {
        if(count > 0)
            count = 99;
    }
    var minutes = date.getMinutes();
    if(count < levels && minutes > 0){
        var langSingle = "date.minute" + (prefix ? "" : ""),
            langMultiple = "date.minutes" + (prefix ? ".prefixed" : "");
        returnString += (count > 0 ?  ', ' : '') + (minutes > 1 ? langFn(langMultiple,[minutes]) : langFn(langSingle,[minutes]));
        count ++;
    } else {
        if(count > 0)
            count = 99;
    }
    var seconds = date.getSeconds();
    if(count < levels && seconds > 0){
        var langSingle = "date.second" + (prefix ? "" : ""),
            langMultiple = "date.seconds" + (prefix ? ".prefixed" : "");
        returnString += (count > 0 ?  ', ' : '') + (seconds > 1 ? langFn(langMultiple,[seconds]) : langFn(langSingle,[seconds]));
        count ++;
    } else {
        if(count > 0)
            count = 99;
    }
    if(prefix){
        if(returnString == ""){
            returnString = langFn("date.now");
        } else if(past)
            returnString = langFn("date.past",[returnString]);
        else
            returnString = langFn("date.future",[returnString]);
    }
    return returnString;
}
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Mur*_*ray 5

我认为您正在寻找与我想要的东西相同的东西。我尝试使用javascript提供的毫秒数来执行此操作,但这些结果在实际的日期世界中不起作用。如果您想要2016年2月1日与2017年1月31日之间的差额,我想要的结果是1年0个月0天。正好一年(假设您将最后一天算作一整天,就像租住公寓一样)。但是,由于日期范围包括a年,因此毫秒方法将为您提供1年0个月零1天。因此,这是我在javascript中用于Adobe表单(您可以命名字段)的代码:(编辑后,我已更正一个错误)

var f1 = this.getField("LeaseExpiration");
var g1 = this.getField("LeaseStart");


var end = f1.value
var begin = g1.value
var e = new Date(end);
var b = new Date(begin);
var bMonth = b.getMonth();
var bYear = b.getFullYear();
var eYear = e.getFullYear();
var eMonth = e.getMonth();
var bDay = b.getDate();
var eDay = e.getDate() + 1;

if ((eMonth == 0)||(eMonth == 2)||(eMonth == 4)|| (eMonth == 6) || (eMonth == 7) ||(eMonth == 9)||(eMonth == 11))

{
var eDays =  31;
}

if ((eMonth == 3)||(eMonth == 5)||(eMonth == 8)|| (eMonth == 10))

{
var eDays = 30;
}

if (eMonth == 1&&((eYear % 4 == 0) && (eYear % 100 != 0)) || (eYear % 400 == 0))
{
var eDays = 29;
}

if (eMonth == 1&&((eYear % 4 != 0) || (eYear % 100 == 0)))
{
var eDays = 28;
}


if ((bMonth == 0)||(bMonth == 2)||(bMonth == 4)|| (bMonth == 6) || (bMonth == 7) ||(bMonth == 9)||(bMonth == 11))

{
var bDays =  31;
}

if ((bMonth == 3)||(bMonth == 5)||(bMonth == 8)|| (bMonth == 10))

{
var bDays = 30;
}

if (bMonth == 1&&((bYear % 4 == 0) && (bYear % 100 != 0)) || (bYear % 400 == 0))
{
var bDays = 29;
}

if (bMonth == 1&&((bYear % 4 != 0) || (bYear % 100 == 0)))
{
var bDays = 28;
}


var FirstMonthDiff = bDays - bDay + 1;


if (eDay - bDay < 0)
{

eMonth = eMonth - 1;
eDay = eDay + eDays;

}

var daysDiff = eDay - bDay;

if(eMonth - bMonth < 0)
{
eYear = eYear - 1;
eMonth = eMonth + 12;
}

var monthDiff = eMonth - bMonth;

var yearDiff = eYear - bYear;

if (daysDiff == eDays)
{
daysDiff = 0;
monthDiff = monthDiff + 1;

if (monthDiff == 12)
{
monthDiff = 0;
yearDiff = yearDiff + 1;
}

}

if ((FirstMonthDiff != bDays)&&(eDay - 1 == eDays))

{
daysDiff = FirstMonthDiff;

}
event.value = yearDiff + " Year(s)" + " " + monthDiff + " month(s) " + daysDiff + " days(s)"
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  • 将下面的代码片段更改为下面的 if (eDay - bDay &lt; 0) { eMonth = eMonth - 1; eDay = eDay + bDays;} (2认同)

Joe*_*ino 0

我个人会使用http://www.datejs.com/,非常方便。具体来说,查看 time.js 文件:http://code.google.com/p/datejs/source/browse/trunk/src/time.js