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

这将显示你过去和以前的时间格式,如“2天前”“从现在开始10分钟”,你可以传递给它一个日期对象,数字时间戳或日期字符串

function time_ago(time) { switch (typeof time) { case 'number': break; case 'string': time = +new Date(time); break; case 'object': if (time.constructor === Date) time = time.getTime(); break; default: time = +new Date(); } var time_formats = [ [60, 'seconds', 1], // 60 [120, '1 minute ago', '1 minute from now'], // 60*2 [3600, 'minutes', 60], // 60*60, 60 [7200, '1 hour ago', '1 hour from now'], // 60*60*2 [86400, 'hours', 3600], // 60*60*24, 60*60 [172800, 'Yesterday', 'Tomorrow'], // 60*60*24*2 [604800, 'days', 86400], // 60*60*24*7, 60*60*24 [1209600, 'Last week', 'Next week'], // 60*60*24*7*4*2 [2419200, 'weeks', 604800], // 60*60*24*7*4, 60*60*24*7 [4838400, 'Last month', 'Next month'], // 60*60*24*7*4*2 [29030400, 'months', 2419200], // 60*60*24*7*4*12, 60*60*24*7*4 [58060800, 'Last year', 'Next year'], // 60*60*24*7*4*12*2 [2903040000, 'years', 29030400], // 60*60*24*7*4*12*100, 60*60*24*7*4*12 [5806080000, 'Last century', 'Next century'], // 60*60*24*7*4*12*100*2 [58060800000, 'centuries', 2903040000] // 60*60*24*7*4*12*100*20, 60*60*24*7*4*12*100 ]; var seconds = (+new Date() - time) / 1000, token = 'ago', list_choice = 1; if (seconds == 0) { return 'Just now' } if (seconds < 0) { seconds = Math.abs(seconds); token = 'from now'; list_choice = 2; } var i = 0, format; while (format = time_formats[i++]) if (seconds < format[0]) { if (typeof format[2] == 'string') return format[list_choice]; else return Math.floor(seconds / format[2]) + ' ' + format[1] + ' ' + token; } return time; } var aDay = 24 * 60 * 60 * 1000; console.log(time_ago(new Date(Date.now() - aDay))); console.log(time_ago(new Date(Date.now() - aDay * 2)));

其他回答

从现在开始,Unix时间戳参数

function timeSince(ts){
    now = new Date();
    ts = new Date(ts*1000);
    var delta = now.getTime() - ts.getTime();

    delta = delta/1000; //us to s

    var ps, pm, ph, pd, min, hou, sec, days;

    if(delta<=59){
        ps = (delta>1) ? "s": "";
        return delta+" second"+ps
    }

    if(delta>=60 && delta<=3599){
        min = Math.floor(delta/60);
        sec = delta-(min*60);
        pm = (min>1) ? "s": "";
        ps = (sec>1) ? "s": "";
        return min+" minute"+pm+" "+sec+" second"+ps;
    }

    if(delta>=3600 && delta<=86399){
        hou = Math.floor(delta/3600);
        min = Math.floor((delta-(hou*3600))/60);
        ph = (hou>1) ? "s": "";
        pm = (min>1) ? "s": "";
        return hou+" hour"+ph+" "+min+" minute"+pm;
    } 

    if(delta>=86400){
        days = Math.floor(delta/86400);
        hou =  Math.floor((delta-(days*86400))/60/60);
        pd = (days>1) ? "s": "";
        ph = (hou>1) ? "s": "";
        return days+" day"+pd+" "+hou+" hour"+ph;
    }

}

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 >

您可能需要查看humanized_time_span: https://github.com/layam/js_humanized_time_span

它与框架无关,并且完全可定制。

只需下载/包含脚本,然后你可以这样做:

humanized_time_span("2011-05-11 12:00:00")  
   => '3 hours ago'

humanized_time_span("2011-05-11 12:00:00", "2011-05-11 16:00:00)  
   => '4 hours ago'

甚至是这样:

var custom_date_formats = {
  past: [
    { ceiling: 60, text: "less than a minute ago" },
    { ceiling: 86400, text: "$hours hours, $minutes minutes and $seconds seconds ago" },
    { ceiling: null, text: "$years years ago" }
  ],
  future: [
    { ceiling: 60, text: "in less than a minute" },
    { ceiling: 86400, text: "in $hours hours, $minutes minutes and $seconds seconds time" },
    { ceiling: null, text: "in $years years" }
  ]
}

humanized_time_span("2010/09/10 10:00:00", "2010/09/10 10:00:05", custom_date_formats) 
  => "less than a minute ago"

更多信息请阅读文档。

回复@Stas Parshin的答案,这是最好的答案,代码更少,但它在与typescript一起使用时有bug, Intl的.format函数需要2个输入

number, Units - i.e of type 'RelativeTimeFormatUnit' so if you pass a object key typescript will through error saying unit must be of type RelativeTimeFormatUnit and not of type string, so the work-around for this is to use the type to make another list of same type and rest you can have look at code... Happy coding. console.log(timeAgo('2021-08-09T15:29:01+0000')); function timeAgo(input) { const date = (input instanceof Date) ? input : new Date(input); const formatter = new Intl.RelativeTimeFormat('en'); const ranges = { years: 3600 * 24 * 365, months: 3600 * 24 * 30, weeks: 3600 * 24 * 7, days: 3600 * 24, hours: 3600, minutes: 60, seconds: 1 }; type RelativeTimeFormatUnit = | "year" | "years" | "quarter" | "quarters" | "month" | "months" | "week" | "weeks" | "day" | "days" | "hour" | "hours" | "minute" | "minutes" | "second" | "seconds" ; const units: RelativeTimeFormatUnit[] = ["years", "months", "weeks", "days", "hours", "minutes", "seconds"]; // order matters here. const secondsElapsed = (date.getTime() - Date.now()) / 1000; for (let key in ranges) { let i = 0; if (ranges[key] < Math.abs(secondsElapsed)) { const delta = secondsElapsed / ranges[key]; return formatter.format(Math.round(delta), units[i++]); } } }

function dateToHowManyAgo(stringDate){
    var currDate = new Date();
    var diffMs=currDate.getTime() - new Date(stringDate).getTime();
    var sec=diffMs/1000;
    if(sec<60)
        return parseInt(sec)+' second'+(parseInt(sec)>1?'s':'')+' ago';
    var min=sec/60;
    if(min<60)
        return parseInt(min)+' minute'+(parseInt(min)>1?'s':'')+' ago';
    var h=min/60;
    if(h<24)
        return parseInt(h)+' hour'+(parseInt(h)>1?'s':'')+' ago';
    var d=h/24;
    if(d<30)
        return parseInt(d)+' day'+(parseInt(d)>1?'s':'')+' ago';
    var m=d/30;
    if(m<12)
        return parseInt(m)+' month'+(parseInt(m)>1?'s':'')+' ago';
    var y=m/12;
    return parseInt(y)+' year'+(parseInt(y)>1?'s':'')+' ago';
}
console.log(dateToHowManyAgo('2019-11-07 19:17:06'));