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
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!)