问题是如何将JavaScript Date格式化为一个字符串,声明时间经过,类似于您在Stack Overflow上看到的时间显示方式。

e.g.

1分钟前 1小时前 1天前 1个月前 一年前


当前回答

你可以使用Luxon:

DateTime.toRelative()

https://moment.github.io/luxon/api-docs/index.html#datetimetorelative

这里还有一些例子:

const DateTime = luxon.DateTime; // toRelative console.log(DateTime.now().minus({ seconds: 1 }).toRelative()) // 1 second ago console.log(DateTime.now().minus({ seconds: 10 }).toRelative()) // 10 seconds ago console.log(DateTime.now().minus({ days: 2 }).toRelative()) console.log(DateTime.now().minus({ years: 3 }).toRelative()) console.log(DateTime.now().plus({ years: 3 }).toRelative()) console.log(DateTime.now().plus({ days: 2 }).toRelative()) console.log(DateTime.now().plus({ seconds: 1 }).toRelative()) console.log(DateTime.now().plus({ seconds: 10 }).toRelative()) // toRelativeCalendar console.log(DateTime.now().minus({ seconds: 1 }).toRelativeCalendar()) // today console.log(DateTime.now().minus({ seconds: 10 }).toRelativeCalendar()) // today console.log(DateTime.now().minus({ days: 2 }).toRelativeCalendar()) console.log(DateTime.now().minus({ years: 3 }).toRelativeCalendar()) console.log(DateTime.now().plus({ years: 3 }).toRelativeCalendar()) console.log(DateTime.now().plus({ days: 2 }).toRelativeCalendar()) console.log(DateTime.now().plus({ seconds: 1 }).toRelativeCalendar()) console.log(DateTime.now().plus({ seconds: 10 }).toRelativeCalendar()) <script src="https://moment.github.io/luxon/global/luxon.min.js"></script>

其他回答

我一直在寻找这个问题的答案,并且几乎实现了其中一个解决方案,但一位同事提醒我检查react-intl库,因为我们已经在使用它了。

所以在解决方案中…在使用react-intl库的情况下,它们有一个<FormattedRelative>组件。

https://github.com/yahoo/react-intl/wiki/Components#formattedrelative

function timeSince(date) { var seconds = Math.floor((new Date() - date) / 1000); var interval = seconds / 31536000; if (interval > 1) { return Math.floor(interval) + " years"; } interval = seconds / 2592000; if (interval > 1) { return Math.floor(interval) + " months"; } interval = seconds / 86400; if (interval > 1) { return Math.floor(interval) + " days"; } interval = seconds / 3600; if (interval > 1) { return Math.floor(interval) + " hours"; } interval = seconds / 60; if (interval > 1) { return Math.floor(interval) + " minutes"; } return Math.floor(seconds) + " seconds"; } var aDay = 24*60*60*1000; console.log(timeSince(new Date(Date.now()-aDay))); console.log(timeSince(new Date(Date.now()-aDay*2)));

以上答案适用于旧的java脚本。但它在新的EC6 JavaScript或TypeScript上运行得不太好。下面是一个非常简短和简单的函数,用于最新的JavaScript, TypeScript, AngularJs, ReactJs和NodeJs,根据给定的日期和时间返回时间。

  public timeAgo(date) {
    var seconds = Math.floor((new Date().getTime() - new Date(date).getTime()) / 1000);
    var interval = seconds / 31536000;
    if (interval > 1) return Math.floor(interval) + " years";
    interval = seconds / 2592000;
    if (interval > 1) return Math.floor(interval) + " months";
    interval = seconds / 86400;
    if (interval > 1) return Math.floor(interval) + " days";
    interval = seconds / 3600;
    if (interval > 1) return Math.floor(interval) + " hours";
    interval = seconds / 60;
    if (interval > 1) return Math.floor(interval) + " minutes";
    return Math.floor(seconds) + " seconds";
  }

console.log(timeAgo('2022-08-12 20:50:20'));
// 2 hours ago, as per the given date time string.

简单易读版本:

const relativeTimePeriods = [
    [31536000, 'year'],
    [2419200, 'month'],
    [604800, 'week'],
    [86400, 'day'],
    [3600, 'hour'],
    [60, 'minute'],
    [1, 'second']
];

function relativeTime(date, isUtc=true) {
    if (!(date instanceof Date)) date = new Date(date * 1000);
    const seconds = (new Date() - date) / 1000;
    for (let [secondsPer, name] of relativeTimePeriods) {
        if (seconds >= secondsPer) {
            const amount = Math.floor(seconds / secondsPer);
            return `${amount} ${name}${amount ? 's' : ''}s ago`;
        }
    }
    return 'Just now';
}

由@user1012181提供的ES6版本代码:

const epochs = [
    ['year', 31536000],
    ['month', 2592000],
    ['day', 86400],
    ['hour', 3600],
    ['minute', 60],
    ['second', 1]
];

const getDuration = (timeAgoInSeconds) => {
    for (let [name, seconds] of epochs) {
        const interval = Math.floor(timeAgoInSeconds / seconds);
        if (interval >= 1) {
            return {
                interval: interval,
                epoch: name
            };
        }
    }
};

const timeAgo = (date) => {
    const timeAgoInSeconds = Math.floor((new Date() - new Date(date)) / 1000);
    const {interval, epoch} = getDuration(timeAgoInSeconds);
    const suffix = interval === 1 ? '' : 's';
    return `${interval} ${epoch}${suffix} ago`;
};

由@ibe-vanmeenen编辑建议。(谢谢!)