javascript - Datejs获取日期之间的差异

所以这是我的目标，我想确定某个日期是多久以前可读的字符串。类似于：“ 4天3小时前”。我正在使用datejs，目前我的两个日期都作为时代。

我知道我可以从另一个减去一个纪元，取其差值，然后手动计算要输出的字符串。但这感觉就像datejs可以处理的。但是，我看不到任何文档（涉及任何js库）来表达将日期对象或时期转换为一定时间的方法。

关于如何最好地处理的任何想法？

更新

让我更加清楚。当我说人类可读的意思是


1天前是“昨天”
2天前是'2天前'
30秒前是“现在”
1年3天12秒钟之前是'1年前'


我最终做了什么

    prettyAgo = (ago) ->
        ago = ago.getTime() if typeof ago is 'object'
        diff = ((new Date()).getTime() - ago) / 1000
        dayDiff = Math.floor(diff / 86400)
        return false if isNaN(dayDiff) or dayDiff < 0

        writer = ''

        if dayDiff is 0
            if diff < 1 then writer             = 'just now'
            else if diff < 60 then writer       = Math.floor(diff)+' seconds ago'
            else if diff < 120 then writer      = '1 minute ago'
            else if diff < 3600 then writer     = Math.floor(diff / 60)+' minutes ago'
            else if diff < 7200 then writer     = '1 hour ago'
            else if diff < 86400 then writer    = Math.floor(diff / 3600)+' hours ago'
        else if dayDiff is 1 then writer        = 'yesterday'
        else if dayDiff < 7 then writer         = dayDiff+' days ago'
        else if dayDiff is 7 then writer        = '1 week ago'
        else if dayDiff < 31 then writer        = Math.ceil( dayDiff / 7 )+' weeks ago'
        else writer                             = new Date(ago).toString 'MMM yyyy'

        return writer

最佳答案

DateJS通过包含一个称为TimeSpan的类（在time.js中定义，而不在基本JS文件中定义）来处理此问题，该类为您进行计算。它不进行格式化，因此您需要这样的内容：

function showDiff(date1, date2) {
    var diff, rv, yrs;

    diff = new TimeSpan(date2 - date1);
    if (diff.days > 0) {
        // A day or more difference
        if (diff.days === 1) {
            rv = "yesterday";
        }
        else if (diff.days >= 365) {
            yrs = diff.days / 365;
            if (yrs === 1) {
                rv = "1 year ago";
            }
            else {
                rv = yrs + " years ago";
            }
        }
        else {
            rv = diff.days + " ago";
        }
    }
    else {
        // Less than 1 day difference
        if (diff.hours > 0) {
            if (diff.hours === 1) {
                rv = "1 hour ago";
            }
            else {
                rv = diff.hours + " hours ago";
            }
        }
        else if (diff.minutes > 0) {
            if (diff.minutes === 1) {
                rv = "1 minute ago";
            }
            else {
                rv = diff.minutes + " minutes ago";
            }
        }
        else {
            rv = "just now";
        }

        return rv;
    }

显然，这并不完美，不能解决leap年和其他所有问题，但这应该可以为您带来灵感。