问题是如何将 JavaScript 格式化Date为一个字符串，说明经过的时间，类似于您在 Stack Overflow 上看到的时间显示方式。
例如
- 1 分钟前
- 1小时前
- 1天前
- 1个月前
- 1年前
问问题
230076 次
31 回答
407
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)));于 2010-07-05T07:45:51.093 回答
155
在这种情况下可能有点矫枉过正，但如果机会显示moment.js真是太棒了！
Moment.js 是一个 javascript 日期时间库，要在这种情况下使用它，您可以：
moment(yourdate).fromNow()
http://momentjs.com/docs/#/displaying/fromnow/
2018 附录：Luxon是一个新的现代图书馆，可能值得一看！
于 2013-04-16T16:50:46.090 回答
72
这将向您显示过去和以前的时间格式，例如“2 天前”“10 分钟后”，您可以将其传递给 Date 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)));于 2012-09-18T10:35:32.083 回答
57
我没有检查过（虽然不难），但我认为Stack Exchange 网站使用jquery.timeago插件来创建这些时间字符串。
该插件非常易于使用，并且干净且自动更新。
这是一个快速示例（来自插件的主页）：
首先，加载 jQuery 和插件：
<script src="jquery.min.js" type="text/javascript"></script> <script src="jquery.timeago.js" type="text/javascript"></script>现在，让我们将它附加到准备好的 DOM 上的时间戳：
jQuery(document).ready(function() {
jQuery("abbr.timeago").timeago(); });这会将标题中
abbr具有类和 ISO 8601 时间戳的所有元素转换为 ： 大约一年前。随着时间的推移，时间戳会自动更新。timeago<abbr class="timeago" title="2008-07-17T09:24:17Z">July 17, 2008</abbr><abbr class="timeago" title="July 17, 2008">about a year ago</abbr>
于 2010-07-05T07:59:05.950 回答
28
这是对 Sky Sander 解决方案的轻微修改，它允许将日期作为字符串输入，并且能够显示像“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)));于 2014-04-24T03:30:12.853 回答
18
Lokely使用的较短版本：
const intervals = [
{ label: 'year', seconds: 31536000 },
{ label: 'month', seconds: 2592000 },
{ label: 'day', seconds: 86400 },
{ label: 'hour', seconds: 3600 },
{ label: 'minute', seconds: 60 },
{ label: 'second', seconds: 1 }
];
function timeSince(date) {
const seconds = Math.floor((Date.now() - date.getTime()) / 1000);
const interval = intervals.find(i => i.seconds < seconds);
const count = Math.floor(seconds / interval.seconds);
return `${count} ${interval.label}${count !== 1 ? 's' : ''} ago`;
}
于 2017-10-29T23:40:48.817 回答
15
您可能想查看humanized_time_span：https ://github.com/layam/js_humanized_time_span
它与框架无关且完全可定制。
只需下载/包含脚本，然后您就可以这样做：
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'
甚至这个：
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 $years years" }
]
}
humanized_time_span("2010/09/10 10:00:00", "2010/09/10 10:00:05", custom_date_formats)
=> "less than a minute ago"
阅读文档以获取更多信息。
于 2011-05-11T14:30:37.523 回答
15
@user1012181 提供的代码的 ES6 版本：
const epochs = [
['year', 31536000],
['month', 2592000],
['day', 86400],
['hour', 3600],
['minute', 60],
['second', 1]
];
const getDuration = (timeAgoInSeconds) => {
for (let [name, seconds] of epochs) {
const interval = Math.floor(timeAgoInSeconds / seconds);
if (interval >= 1) {
return {
interval: interval,
epoch: name
};
}
}
};
const timeAgo = (date) => {
const timeAgoInSeconds = Math.floor((new Date() - new Date(date)) / 1000);
const {interval, epoch} = getDuration(timeAgoInSeconds);
const suffix = interval === 1 ? '' : 's';
return `${interval} ${epoch}${suffix} ago`;
};
使用@ibe-vanmeenen 建议进行编辑。（谢谢！）
于 2015-11-02T21:28:55.190 回答
14
将上面的功能更改为
function timeSince(date) {
var seconds = Math.floor(((new Date().getTime()/1000) - date)),
interval = Math.floor(seconds / 31536000);
if (interval > 1) return interval + "y";
interval = Math.floor(seconds / 2592000);
if (interval > 1) return interval + "m";
interval = Math.floor(seconds / 86400);
if (interval >= 1) return interval + "d";
interval = Math.floor(seconds / 3600);
if (interval >= 1) return interval + "h";
interval = Math.floor(seconds / 60);
if (interval > 1) return interval + "m ";
return Math.floor(seconds) + "s";
}
否则它将显示诸如“75 分钟”（1 到 2 小时之间）之类的内容。它现在还假设输入日期是 Unix 时间戳。
于 2011-11-23T18:13:53.600 回答
12
所以这是我的版本，它适用于过去和未来的日期。它使用Intl.RelativeTimeFormat提供本地化字符串，而不是硬编码字符串。您可以将日期作为时间戳、日期对象或可解析的日期字符串传递。
/**
* Human readable elapsed or remaining time (example: 3 minutes ago)
* @param {Date|Number|String} date A Date object, timestamp or string parsable with Date.parse()
* @param {Date|Number|String} [nowDate] A Date object, timestamp or string parsable with Date.parse()
* @param {Intl.RelativeTimeFormat} [trf] A Intl formater
* @return {string} Human readable elapsed or remaining time
* @author github.com/victornpb
* @see https://stackoverflow.com/a/67338038/938822
*/
function fromNow(date, nowDate = Date.now(), rft = new Intl.RelativeTimeFormat(undefined, { numeric: "auto" })) {
const SECOND = 1000;
const MINUTE = 60 * SECOND;
const HOUR = 60 * MINUTE;
const DAY = 24 * HOUR;
const WEEK = 7 * DAY;
const MONTH = 30 * DAY;
const YEAR = 365 * DAY;
const intervals = [
{ ge: YEAR, divisor: YEAR, unit: 'year' },
{ ge: MONTH, divisor: MONTH, unit: 'month' },
{ ge: WEEK, divisor: WEEK, unit: 'week' },
{ ge: DAY, divisor: DAY, unit: 'day' },
{ ge: HOUR, divisor: HOUR, unit: 'hour' },
{ ge: MINUTE, divisor: MINUTE, unit: 'minute' },
{ ge: 30 * SECOND, divisor: SECOND, unit: 'seconds' },
{ ge: 0, divisor: 1, text: 'just now' },
];
const now = typeof nowDate === 'object' ? nowDate.getTime() : new Date(nowDate).getTime();
const diff = now - (typeof date === 'object' ? date : new Date(date)).getTime();
const diffAbs = Math.abs(diff);
for (const interval of intervals) {
if (diffAbs >= interval.ge) {
const x = Math.round(Math.abs(diff) / interval.divisor);
const isFuture = diff < 0;
return interval.unit ? rft.format(isFuture ? x : -x, interval.unit) : interval.text;
}
}
}// examples
fromNow('2020-01-01') // 9 months ago
fromNow(161651684156) // 4 days ago
fromNow(new Date()-1) // just now
fromNow(30000 + Date.now()) // in 30 seconds
fromNow(Date.now() + (1000*60*60*24)) // in 1 day
fromNow(new Date('2029-12-01Z00:00:00.000')) // in 9 years
不使用 Intl.RelativeTimeFormat 的替代方案
/**
* Human readable elapsed or remaining time (example: 3 minutes ago)
* @param {Date|Number|String} date A Date object, timestamp or string parsable with Date.parse()
* @return {string} Human readable elapsed or remaining time
* @author github.com/victornpb
* @see https://stackoverflow.com/a/67338038/938822
*/
function fromNow(date) {
const SECOND = 1000;
const MINUTE = 60 * SECOND;
const HOUR = 60 * MINUTE;
const DAY = 24 * HOUR;
const WEEK = 7 * DAY;
const MONTH = 30 * DAY;
const YEAR = 365 * DAY;
const units = [
{ max: 30 * SECOND, divisor: 1, past1: 'just now', pastN: 'just now', future1: 'just now', futureN: 'just now' },
{ max: MINUTE, divisor: SECOND, past1: 'a second ago', pastN: '# seconds ago', future1: 'in a second', futureN: 'in # seconds' },
{ max: HOUR, divisor: MINUTE, past1: 'a minute ago', pastN: '# minutes ago', future1: 'in a minute', futureN: 'in # minutes' },
{ max: DAY, divisor: HOUR, past1: 'an hour ago', pastN: '# hours ago', future1: 'in an hour', futureN: 'in # hours' },
{ max: WEEK, divisor: DAY, past1: 'yesterday', pastN: '# days ago', future1: 'tomorrow', futureN: 'in # days' },
{ max: 4 * WEEK, divisor: WEEK, past1: 'last week', pastN: '# weeks ago', future1: 'in a week', futureN: 'in # weeks' },
{ max: YEAR, divisor: MONTH, past1: 'last month', pastN: '# months ago', future1: 'in a month', futureN: 'in # months' },
{ max: 100 * YEAR, divisor: YEAR, past1: 'last year', pastN: '# years ago', future1: 'in a year', futureN: 'in # years' },
{ max: 1000 * YEAR, divisor: 100 * YEAR, past1: 'last century', pastN: '# centuries ago', future1: 'in a century', futureN: 'in # centuries' },
{ max: Infinity, divisor: 1000 * YEAR, past1: 'last millennium', pastN: '# millennia ago', future1: 'in a millennium', futureN: 'in # millennia' },
];
const diff = Date.now() - (typeof date === 'object' ? date : new Date(date)).getTime();
const diffAbs = Math.abs(diff);
for (const unit of units) {
if (diffAbs < unit.max) {
const isFuture = diff < 0;
const x = Math.round(Math.abs(diff) / unit.divisor);
if (x <= 1) return isFuture ? unit.future1 : unit.past1;
return (isFuture ? unit.futureN : unit.pastN).replace('#', x);
}
}
};于 2021-04-30T17:21:23.447 回答
11
可读性强且跨浏览器兼容的代码：
正如@Travis 给出的
var DURATION_IN_SECONDS = {
epochs: ['year', 'month', 'day', 'hour', 'minute'],
year: 31536000,
month: 2592000,
day: 86400,
hour: 3600,
minute: 60
};
function getDuration(seconds) {
var epoch, interval;
for (var i = 0; i < DURATION_IN_SECONDS.epochs.length; i++) {
epoch = DURATION_IN_SECONDS.epochs[i];
interval = Math.floor(seconds / DURATION_IN_SECONDS[epoch]);
if (interval >= 1) {
return {
interval: interval,
epoch: epoch
};
}
}
};
function timeSince(date) {
var seconds = Math.floor((new Date() - new Date(date)) / 1000);
var duration = getDuration(seconds);
var suffix = (duration.interval > 1 || duration.interval === 0) ? 's' : '';
return duration.interval + ' ' + duration.epoch + suffix;
};
alert(timeSince('2015-09-17T18:53:23'));于 2015-09-17T19:10:01.870 回答
8
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;
}
}
于 2012-11-10T15:03:46.823 回答
7
这应该正确处理任何有效的时间戳，包括 Date.now()、单数单位和未来日期。我遗漏了几个月，但这些应该很容易添加。我尽量保持它的可读性。
function getTimeInterval(date) {
let seconds = Math.floor((Date.now() - date) / 1000);
let unit = "second";
let direction = "ago";
if (seconds < 0) {
seconds = -seconds;
direction = "from now";
}
let value = seconds;
if (seconds >= 31536000) {
value = Math.floor(seconds / 31536000);
unit = "year";
} else if (seconds >= 86400) {
value = Math.floor(seconds / 86400);
unit = "day";
} else if (seconds >= 3600) {
value = Math.floor(seconds / 3600);
unit = "hour";
} else if (seconds >= 60) {
value = Math.floor(seconds / 60);
unit = "minute";
}
if (value != 1)
unit = unit + "s";
return value + " " + unit + " " + direction;
}
console.log(getTimeInterval(Date.now())); // 0 seconds ago
console.log(getTimeInterval(Date.now() + 1000)); // 1 second from now
console.log(getTimeInterval(Date.now() - 1000)); // 1 second ago
console.log(getTimeInterval(Date.now() + 60000)); // 1 minute from now
console.log(getTimeInterval(Date.now() - 120000)); // 2 minutes ago
console.log(getTimeInterval(Date.now() + 120000)); // 2 minutes from now
console.log(getTimeInterval(Date.now() + 3600000)); // 1 hour from now
console.log(getTimeInterval(Date.now() + 360000000000)); // 11 years from now
console.log(getTimeInterval(0)); // 49 years ago于 2019-11-22T23:58:58.903 回答
6
另一种看法Intl.RelativeTimeFormat
- 支持过去和未来的日期
- 接受
String和Date - 自定义范围很容易添加（编辑
ranges） - 可以很容易地翻译
Intl.RelativeTimeFormat('ua')
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
};
const secondsElapsed = (date.getTime() - Date.now()) / 1000;
for (let key in ranges) {
if (ranges[key] < Math.abs(secondsElapsed)) {
const delta = secondsElapsed / ranges[key];
return formatter.format(Math.round(delta), key);
}
}
}于 2021-09-09T18:04:29.220 回答
5
也可以使用 dayjs 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
于 2019-07-02T18:35:14.773 回答
5
尽管这个问题是很久以前提出的，但写下这个答案希望它会对某人有所帮助。
通过您要开始计数的日期。momentjsmoment().fromNow()的使用：（在此处查看更多信息）
getRelativeTime(date) {
const d = new Date(date * 1000);
return moment(d).fromNow();
}
如果您想更改从现在开始的日期提供的信息，您可以写下您的自定义相对时间。
例如，在我自己的情况下，我想打印'one month ago'而不是'a month ago'（由 moment(d).fromNow() 提供）。在这种情况下，您可以编写下面给出的内容。
moment.updateLocale('en', {
relativeTime: {
future: 'in %s',
past: '%s ago',
s: 'a few seconds',
ss: '%d seconds',
m: '1 m',
mm: '%d minutes',
h: '1 h',
hh: '%d hours',
d: '1 d',
dd: '%d days',
M: '1 month',
MM: '%d months',
y: '1 y',
yy: '%d years'
}
});
注意：我在Angular 6中为项目编写了代码
于 2018-09-10T10:52:32.900 回答
4
简单易读的版本：
const NOW = new Date()
const times = [["second", 1], ["minute", 60], ["hour", 3600], ["day", 86400], ["week", 604800], ["month", 2592000], ["year", 31536000]]
function timeAgo(date) {
var diff = Math.round((NOW - date) / 1000)
for (var t = 0; t < times.length; t++) {
if (diff < times[t][1]) {
if (t == 0) {
return "Just now"
} else {
diff = Math.round(diff / times[t - 1][1])
return diff + " " + times[t - 1][0] + (diff == 1?" ago":"s ago")
}
}
}
}
于 2018-10-15T06:28:54.733 回答
3
我用 js 和 python 写了一个，在两个项目中使用，非常漂亮和简单：一个简单的库（小于 2kb），用于格式化日期和*** time ago语句。
简单、小巧、易于使用且经过良好测试。
npm install timeago.jsimport timeago from 'timeago.js'; // or use script tag使用 api
format。
样本：
var timeagoIns = timeago();
timeagoIns .format('2016-06-12');
您也可以实时渲染。
var timeagoIns = timeago();
timeagoIns.render(document.querySelectorAll('time'));
于 2016-06-24T12:52:17.550 回答
3
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'));
于 2019-11-07T19:50:47.357 回答
2
我修改了 Sky Sanders 的版本。Math.floor(...) 操作在 if 块中进行评估
var timeSince = function(date) {
var seconds = Math.floor((new Date() - date) / 1000);
var months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"];
if (seconds < 5){
return "just now";
}else if (seconds < 60){
return seconds + " seconds ago";
}
else if (seconds < 3600) {
minutes = Math.floor(seconds/60)
if(minutes > 1)
return minutes + " minutes ago";
else
return "1 minute ago";
}
else if (seconds < 86400) {
hours = Math.floor(seconds/3600)
if(hours > 1)
return hours + " hours ago";
else
return "1 hour ago";
}
//2 days and no more
else if (seconds < 172800) {
days = Math.floor(seconds/86400)
if(days > 1)
return days + " days ago";
else
return "1 day ago";
}
else{
//return new Date(time).toLocaleDateString();
return date.getDate().toString() + " " + months[date.getMonth()] + ", " + date.getFullYear();
}
}
于 2016-01-20T02:02:48.857 回答
2
回答10年前的问题以帮助新人。
我们可以将这个包用于那个javascript-time-ago
// Load locale-specific relative date/time formatting rules.
import en from 'javascript-time-ago/locale/en'
// Add locale-specific relative date/time formatting rules.
TimeAgo.addLocale(en)
// Create relative date/time formatter.
const timeAgo = new TimeAgo('en-US')
timeAgo.format(new Date())
// "just now"
timeAgo.format(Date.now() - 60 * 1000)
// "a minute ago"
timeAgo.format(Date.now() - 2 * 60 * 60 * 1000)
// "2 hours ago"
timeAgo.format(Date.now() - 24 * 60 * 60 * 1000)
// "a day ago"
于 2020-09-10T05:52:47.697 回答
2
我根据其他答案对此进行了尝试。
function timeSince(date) {
let minute = 60;
let hour = minute * 60;
let day = hour * 24;
let month = day * 30;
let year = day * 365;
let suffix = ' ago';
let elapsed = Math.floor((Date.now() - date) / 1000);
if (elapsed < minute) {
return 'just now';
}
// get an array in the form of [number, string]
let a = elapsed < hour && [Math.floor(elapsed / minute), 'minute'] ||
elapsed < day && [Math.floor(elapsed / hour), 'hour'] ||
elapsed < month && [Math.floor(elapsed / day), 'day'] ||
elapsed < year && [Math.floor(elapsed / month), 'month'] ||
[Math.floor(elapsed / year), 'year'];
// pluralise and append suffix
return a[0] + ' ' + a[1] + (a[0] === 1 ? '' : 's') + suffix;
}
于 2019-10-13T12:01:53.190 回答
2
const createdAt = moment(created_at).fromNow()
和定制的解决方案
const duration = moment.duration(moment().diff(moment(created_at)))
const createdAt = duration.as('week') >= 1
? `${Math.floor(duration.as('week'))} week(s)`
: duration.as('day') >= 1
? `${Math.floor(duration.as('day'))} day(s)`
: duration.as('hour') >= 1
? `${Math.floor(duration.as('hour'))} hour(s)`
: `${Math.floor(duration.as('minute'))} minute(s)`
于 2021-11-18T07:12:21.480 回答
1
function timeago(date) {
var seconds = Math.floor((new Date() - date) / 1000);
if(Math.round(seconds/(60*60*24*365.25)) >= 2) return Math.round(seconds/(60*60*24*365.25)) + " years ago";
else if(Math.round(seconds/(60*60*24*365.25)) >= 1) return "1 year ago";
else if(Math.round(seconds/(60*60*24*30.4)) >= 2) return Math.round(seconds/(60*60*24*30.4)) + " months ago";
else if(Math.round(seconds/(60*60*24*30.4)) >= 1) return "1 month ago";
else if(Math.round(seconds/(60*60*24*7)) >= 2) return Math.round(seconds/(60*60*24*7)) + " weeks ago";
else if(Math.round(seconds/(60*60*24*7)) >= 1) return "1 week ago";
else if(Math.round(seconds/(60*60*24)) >= 2) return Math.round(seconds/(60*60*24)) + " days ago";
else if(Math.round(seconds/(60*60*24)) >= 1) return "1 day ago";
else if(Math.round(seconds/(60*60)) >= 2) return Math.round(seconds/(60*60)) + " hours ago";
else if(Math.round(seconds/(60*60)) >= 1) return "1 hour ago";
else if(Math.round(seconds/60) >= 2) return Math.round(seconds/60) + " minutes ago";
else if(Math.round(seconds/60) >= 1) return "1 minute ago";
else if(seconds >= 2)return seconds + " seconds ago";
else return seconds + "1 second ago";
}
于 2017-04-01T19:16:58.917 回答
1
我的解决方案..
(function(global){
const SECOND = 1;
const MINUTE = 60;
const HOUR = 3600;
const DAY = 86400;
const MONTH = 2629746;
const YEAR = 31556952;
const DECADE = 315569520;
global.timeAgo = function(date){
var now = new Date();
var diff = Math.round(( now - date ) / 1000);
var unit = '';
var num = 0;
var plural = false;
switch(true){
case diff <= 0:
return 'just now';
break;
case diff < MINUTE:
num = Math.round(diff / SECOND);
unit = 'sec';
plural = num > 1;
break;
case diff < HOUR:
num = Math.round(diff / MINUTE);
unit = 'min';
plural = num > 1;
break;
case diff < DAY:
num = Math.round(diff / HOUR);
unit = 'hour';
plural = num > 1;
break;
case diff < MONTH:
num = Math.round(diff / DAY);
unit = 'day';
plural = num > 1;
break;
case diff < YEAR:
num = Math.round(diff / MONTH);
unit = 'month';
plural = num > 1;
break;
case diff < DECADE:
num = Math.round(diff / YEAR);
unit = 'year';
plural = num > 1;
break;
default:
num = Math.round(diff / YEAR);
unit = 'year';
plural = num > 1;
}
var str = '';
if(num){
str += `${num} `;
}
str += `${unit}`;
if(plural){
str += 's';
}
str += ' ago';
return str;
}
})(window);
console.log(timeAgo(new Date()));
console.log(timeAgo(new Date('Jun 03 2018 15:12:19 GMT+0300 (FLE Daylight Time)')));
console.log(timeAgo(new Date('Jun 03 2018 13:12:19 GMT+0300 (FLE Daylight Time)')));
console.log(timeAgo(new Date('May 28 2018 13:12:19 GMT+0300 (FLE Daylight Time)')));
console.log(timeAgo(new Date('May 28 2017 13:12:19 GMT+0300 (FLE Daylight Time)')));
console.log(timeAgo(new Date('May 28 2000 13:12:19 GMT+0300 (FLE Daylight Time)')));
console.log(timeAgo(new Date('Sep 10 1994 13:12:19 GMT+0300 (FLE Daylight Time)')));
于 2018-06-03T12:26:11.840 回答
0
I achieve this by following method
timeAgo = (date) => {
var ms = (new Date()).getTime() - date.getTime();
var seconds = Math.floor(ms / 1000);
var minutes = Math.floor(seconds / 60);
var hours = Math.floor(minutes / 60);
var days = Math.floor(hours / 24);
var months = Math.floor(days / 30);
var years = Math.floor(months / 12);
if (ms === 0) {
return 'Just now';
} if (seconds < 60) {
return seconds + ' seconds Ago';
} if (minutes < 60) {
return minutes + ' minutes Ago';
} if (hours < 24) {
return hours + ' hours Ago';
} if (days < 30) {
return days + ' days Ago';
} if (months < 12) {
return months + ' months Ago';
} else {
return years + ' years Ago';
}
}
console.log(timeAgo(new Date()));
console.log(timeAgo(new Date('Jun 27 2020 10:12:19')));
console.log(timeAgo(new Date('Jun 27 2020 00:12:19')));
console.log(timeAgo(new Date('May 28 2020 13:12:19')));
console.log(timeAgo(new Date('May 28 2017 13:12:19')));
于 2020-07-12T01:09:03.533 回答
0
function calDateAgo(dString=null){
//var dString = "2021-04-1 12:00:00";
var d1 = new Date(dString);
var d2 = new Date();
var t2 = d2.getTime();
var t1 = d1.getTime();
var d1Y = d1.getFullYear();
var d2Y = d2.getFullYear();
var d1M = d1.getMonth();
var d2M = d2.getMonth();
var time_obj = {};
time_obj.year = d2.getFullYear()-d1.getFullYear();
time_obj.month = (d2M+12*d2Y)-(d1M+12*d1Y);
time_obj.week = parseInt((t2-t1)/(24*3600*1000*7));
time_obj.day = parseInt((t2-t1)/(24*3600*1000));
time_obj.hour = parseInt((t2-t1)/(3600*1000));
time_obj.minute = parseInt((t2-t1)/(60*1000));
time_obj.second = parseInt((t2-t1)/(1000));
for (const obj_key in time_obj) {
if(time_obj[obj_key] == 0){
delete time_obj[obj_key];
}
}
var ago_text = 'just now';
if(typeof Object.keys(time_obj)[0] != 'undefined'){
var time_key = Object.keys(time_obj)[0];
var time_val = time_obj[Object.keys(time_obj)[0]];
time_key += (time_val > 1) ? 's':'';
ago_text = time_val+' '+time_key+' ago';
}
return ago_text;
}
于 2021-04-01T08:41:09.267 回答
0
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';
}
于 2021-10-28T12:06:58.627 回答
0
这是@sky-sanders 答案的简化版本。
function timeSince(date) {
var seconds = Math.floor((new Date() - date) / 1000);
var divisors = [31536000, 2592000, 86400, 3600, 60, 1]
var description = ["years", "months", "days", "hours", "minutes", "seconds"]
var result = [];
var interval = seconds;
for (i = 0; i < divisors.length; i++) {
interval = Math.floor(seconds / divisors[i])
if (interval > 1) {
result.push(interval + " " + description[i])
}
seconds -= interval * divisors[i]
}
return result.join(" ")
}
于 2021-03-10T20:46:24.217 回答
0
我一直在寻找这个问题的答案并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。
因此，添加到解决方案中......如果您使用该react-intl库，他们有一个<FormattedRelative>组件。
https://github.com/yahoo/react-intl/wiki/Components#formattedrelative
于 2016-07-13T22:11:49.310 回答
0
这是我所做的（对象返回时间单位及其值）：
function timeSince(post_date, reference)
{
var reference = reference ? new Date(reference) : new Date(),
diff = reference - new Date(post_date + ' GMT-0000'),
date = new Date(diff),
object = { unit: null, value: null };
if (diff < 86400000)
{
var secs = date.getSeconds(),
mins = date.getMinutes(),
hours = date.getHours(),
array = [ ['second', secs], ['minute', mins], ['hour', hours] ];
}
else
{
var days = date.getDate(),
weeks = Math.floor(days / 7),
months = date.getMonth(),
years = date.getFullYear() - 1970,
array = [ ['day', days], ['week', weeks], ['month', months], ['year', years] ];
}
for (var i = 0; i < array.length; i++)
{
array[i][0] += array[i][1] != 1 ? 's' : '';
object.unit = array[i][1] >= 1 ? array[i][0] : object.unit;
object.value = array[i][1] >= 1 ? array[i][1] : object.value;
}
return object;
}于 2017-02-16T02:56:33.353 回答