获取两个日期时间之间的时间差

Question

获取两个日期时间之间的时间差

282

如何计算两个时间之间的差异？例如：

var now  = "04/09/2013 15:00:00";
var then = "04/09/2013 14:20:30";

//expected result:
"00:39:30"

我试过了。

var now = moment("04/09/2013 15:00:00");
var then = moment("04/09/2013 14:20:30");

console.log(moment(moment.duration(now.diff(then))).format("hh:mm:ss"))
//outputs 10:39:30

那里的"10"是什么意思？我现在是在UTC-0300时区。执行moment.duration(now.diff(then))的结果是一个内部值正确的持续时间。

 days: 0
 hours: 0
 milliseconds: 0
 minutes: 39
 months: 0
 seconds: 30
 years: 0

如何将momentjs的持续时间转换为时间间隔？我可以使用：

duration.get("hours") +":"+ duration.get("minutes") +:+ duration.get("seconds")

但是有没有更优雅的东西呢？现在是：

Tue Apr 09 2013 15:00:00 GMT-0300 (E. South America Standard Time)…}

而 moment(moment.duration(now.diff(then))) 是：

Wed Dec 31 1969 22:39:30 GMT-0200 (E. South America Daylight Time)…}

数值为-0200，因为在1969年12月31日使用了夏令时。

- Leo

4

“但我确定我不喜欢约会” :P 你有没有读过这个：http://nodatime.org/unstable/userguide/trivia.html？ - Boris Callens

22个回答

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- Fernando Rossato · Answer 1

在 ES8 中使用 moment，now 和 start 都是 moment 对象。

const duration = moment.duration(now.diff(start));
const timespan = duration.get("hours").toString().padStart(2, '0') +":"+ duration.get("minutes").toString().padStart(2, '0') +":"+ duration.get("seconds").toString().padStart(2, '0');

- skantus · Answer 2

Typescript：以下代码应该可以正常运行，

export const getTimeBetweenDates = ({
  until,
  format
}: {
  until: number;
  format: 'seconds' | 'minutes' | 'hours' | 'days';
}): number => {
  const date = new Date();
  const remainingTime = new Date(until * 1000);
  const getFrom = moment([date.getUTCFullYear(), date.getUTCMonth(), date.getUTCDate()]);
  const getUntil = moment([remainingTime.getUTCFullYear(), remainingTime.getUTCMonth(), remainingTime.getUTCDate()]);
  const diff = getUntil.diff(getFrom, format);
  return !isNaN(diff) ? diff : null;
};

- Dương Đạt · Answer 3

这将返回最大时间差，例如（4秒，2分钟，1小时，2天，3周，4个月，5年）。我用它来通知最近的时间。

function dateDiff(startDate, endDate) {
    let arrDate = ["seconds", "minutes", "hours", "days", "weeks", "months", "years"];
    let dateMap = arrDate.map(e => moment(endDate).diff(startDate, e));
    let index = 6 - dateMap.filter(e => e == 0).length;
    return {
        type: arrDate[index] ?? "seconds",
        value: dateMap[index] ?? 0
    };
}

例子：

dateDiff("2021-06-09 01:00:00", "2021-06-09 04:01:01")

{type: "hours", value: 3}

dateDiff("2021-06-09 01:00:00", "2021-06-12 04:01:01")

{type: "days", value: 3}

dateDiff("2021-06-09 01:00:00", "2021-06-09 01:00:10")

{type: "seconds", value: 10}

- Erik Martín Jordán · Answer 4

以下方法适用于所有情况（日期之间的差异小于24小时和大于24小时的差异）：

// Defining start and end variables
let start = moment('04/09/2013 15:00:00', 'DD/MM/YYYY hh:mm:ss');
let end   = moment('04/09/2013 14:20:30', 'DD/MM/YYYY hh:mm:ss');

// Getting the difference: hours (h), minutes (m) and seconds (s)
let h  = end.diff(start, 'hours');
let m  = end.diff(start, 'minutes') - (60 * h);
let s  = end.diff(start, 'seconds') - (60 * 60 * h) - (60 * m);

// Formating in hh:mm:ss (appends a left zero when num < 10)
let hh = ('0' + h).slice(-2);
let mm = ('0' + m).slice(-2);
let ss = ('0' + s).slice(-2);

console.log(`${hh}:${mm}:${ss}`); // 00:39:30

- Anandan K · Answer 5

日期时间输入

    var dt1 = new Date("2019-1-8 11:19:16");
    var dt2 = new Date("2019-1-8 11:24:16");


    var diff =(dt2.getTime() - dt1.getTime()) ;
    var hours = Math.floor(diff / (1000 * 60 * 60));
    diff -= hours * (1000 * 60 * 60);
    var mins = Math.floor(diff / (1000 * 60));
    diff -= mins * (1000 * 60);


    var response = {
        status : 200,
        Hour : hours,
        Mins : mins
    }

OUTPUT

{
"status": 200,
"Hour": 0,
"Mins": 5
}

- Sanket Parchande · Answer 6

上班时间=06:38，下班时间=15:40

 calTimeDifference(){
        this.start = dailyattendance.InTime.split(":");
        this.end = dailyattendance.OutTime.split(":");
        var time1 = ((parseInt(this.start[0]) * 60) + parseInt(this.start[1]))
        var time2 = ((parseInt(this.end[0]) * 60) + parseInt(this.end[1]));
        var time3 = ((time2 - time1) / 60);
        var timeHr = parseInt(""+time3);
        var  timeMin = ((time2 - time1) % 60);
    }

- Serkan KONAKCI · Answer 7

我使用 TypeScript 创建了一个简单的函数。

const diffDuration: moment.Duration = moment.duration(moment('2017-09-04 12:55').diff(moment('2017-09-02 13:26')));
setDiffTimeString(diffDuration);

function setDiffTimeString(diffDuration: moment.Duration) {
  const str = [];
  diffDuration.years() > 0 ? str.push(`${diffDuration.years()} year(s)`) : null;
  diffDuration.months() > 0 ? str.push(`${diffDuration.months()} month(s)`) : null;
  diffDuration.days() > 0 ? str.push(`${diffDuration.days()} day(s)`) : null;
  diffDuration.hours() > 0 ? str.push(`${diffDuration.hours()} hour(s)`) : null;
  diffDuration.minutes() > 0 ? str.push(`${diffDuration.minutes()} minute(s)`) : null;
  console.log(str.join(', '));
} 
// output: 1 day(s), 23 hour(s), 29 minute(s)

生成 JavaScript 的代码：https://www.typescriptlang.org/play/index.html。

- Ajmal Hasan · Answer 8

使用moment非常简单，下面的代码将返回当前时间与指定时间之间的小时差：

moment().diff('2021-02-17T14:03:55.811000Z', "h")

- Uzma · Answer 9

const getRemainingTime = (t2) => {
  const t1 = new Date().getTime();
  let ts = (t1-t2.getTime()) / 1000;

  var d = Math.floor(ts / (3600*24));
  var h = Math.floor(ts % (3600*24) / 3600);
  var m = Math.floor(ts % 3600 / 60);
  var s = Math.floor(ts % 60);

  console.log(d, h, m, s)

}

- aakarsh · Answer 10

使用Moment.js计算时间戳差异:

获取两个时间戳之间的差异:

Syntax: moment.duration(moment(moment(date1).diff(moment(date2)))).asHours()

小时差异： moment.duration(moment(moment(1590597744551).diff(moment(1590597909877)))).asHours()

分钟差异： moment.duration(moment(moment(1590597744551).diff(moment(1590597909877)))).asMinutes().toFixed()

注意：如果您需要精确值，可以删除.toFixed()。

代码：

const moment = require('moment')

console.log('Date 1',moment(1590597909877).toISOString())
console.log('Date 2',moment(1590597744551).toISOString())
console.log('Date1 - Date 2 time diffrence is : ',moment.duration(moment(moment(1590597909877).diff(moment(1590597744551)))).asMinutes().toFixed()+' minutes')

请参考以下工作示例： https://repl.it/repls/MoccasinDearDimension