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

function mdiv(dividend, divisor) {
    return [ Math.floor(dividend/divisor), dividend % divisor ];
}
// pass in milliseconds, gained by Date1.getTime() - Date2.getTime()
// if max_units is two, the result will be, for example
// 2years 12months ago, or 2hours 38minutes ago
// return formated period

function readable_period(ms, max_units=2){
    let [yy, yr] = mdiv(ms, 3.154e10);
    let [mm, mr] = mdiv(yr, 2.628e9);
    let [dd, dr] = mdiv(mr, 8.64e7);
    let [hh, hr] = mdiv(dr, 3.6e6);
    let [tt, ss] = mdiv(hr, 6e4);

    var ymdht = ['year', 'month', 'day', 'hour', 'minute'];
    let res = [];
    [yy, mm, dd, hh, tt].forEach((tis, ii)=>{
        if(res.length === max_units){return};
        if(tis !== 0){
            res.push(tis === 1 ? `${tis}${ymdht[ii]}` : `${tis}${ymdht[ii]}s`);
        }
    });
    return res.length === 0 ? '' : res.join(' ') + ' ago';
}

其他回答

以下是对Sky Sander的解决方案的轻微修改,允许日期作为字符串输入,并能够显示像“1分钟”而不是“73秒”这样的跨度

var timeSince = function(date) { if (typeof date !== 'object') { date = new Date(date); } var seconds = Math.floor((new Date() - date) / 1000); var intervalType; var interval = Math.floor(seconds / 31536000); if (interval >= 1) { intervalType = 'year'; } else { interval = Math.floor(seconds / 2592000); if (interval >= 1) { intervalType = 'month'; } else { interval = Math.floor(seconds / 86400); if (interval >= 1) { intervalType = 'day'; } else { interval = Math.floor(seconds / 3600); if (interval >= 1) { intervalType = "hour"; } else { interval = Math.floor(seconds / 60); if (interval >= 1) { intervalType = "minute"; } else { interval = seconds; intervalType = "second"; } } } } } if (interval > 1 || interval === 0) { intervalType += 's'; } return interval + ' ' + intervalType; }; var aDay = 24 * 60 * 60 * 1000; console.log(timeSince(new Date(Date.now() - aDay))); console.log(timeSince(new Date(Date.now() - aDay * 2)));

我使用这个包:javascript-time-ago

设置TimeAgo 导入TimeAgo 从javascript-time-ago/locale/en.json导入en TimeAgo.addDefaultLocale (en) 写一个函数 // twitter-now是twitter风格。查看文档以获得更多选项 const clockToDateString = (timestamp) => timeAgo。format(new Date(timestamp.toNumber() * 1000), 'twitter-now') 在dom中使用它 < div > {clockToDateString (post.postTime)} < / div >

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.

也可以使用dayjs的relativeTime插件来解决这个问题。

import * as dayjs from 'dayjs';
import * as relativeTime from 'dayjs/plugin/relativeTime';

dayjs.extend(relativeTime);
dayjs(dayjs('1990')).fromNow(); // x years ago

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)));