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
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 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'));
其他回答
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(少于2kb),用于*** time ago语句格式化日期。
简单,小巧,易于使用,并且经过良好测试。
NPM安装timeago.js 从timeago.js导入timeago;//或使用脚本标签 使用API格式。
示例:
var timeagoIns = timeago();
timeagoIns .format('2016-06-12');
你也可以实时渲染。
var timeagoIns = timeago();
timeagoIns.render(document.querySelectorAll('time'));
这是对国际的另一种看法。RelativeTimeFormat
支持过去日期和将来日期 同时接受字符串和日期 易于添加自定义范围(编辑范围) 可以很容易地翻译为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); } } }
https://jsfiddle.net/tv9701uf
我修改了Sky Sanders的版本。Math.floor(…)操作在if块中计算
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();
}
}
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)));
我的尝试是基于其他的答案。
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;
}