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
将上面的函数更改为
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";
}
否则它会显示“75分钟”(介于1到2小时之间)。它现在还假定输入日期是Unix时间戳。
其他回答
下面是我所做的(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; }
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'));
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)));
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";
}
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!)