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

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);

其他回答

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。

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>

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);

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)
             )
        ));
})();

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;
    }
}