javascript - Datejs 获取日期之间的差异

Question

所以这就是我的目标，我想得到一个特定日期是多久以前的一个非常可读的字符串。类似于:“4 天零 3 小时前”。我正在使用 datejs 并将我的两个日期当前作为纪元。

我知道我可以从另一个纪元中减去一个纪元，取差值，然后手动计算要输出的字符串。但这感觉像是 datejs 会处理的事情。但是，我没有看到任何文档(与此相关的任何 js 库)表达了将日期对象或纪元转换为时间量的方法。

关于如何最好地处理这个问题有什么想法吗？

更新

让我更清楚一点。当我说人类可读时，这意味着

• 1 天前是“昨天”
• 2 天前是“2 天前”
• 30 秒前是“刚刚”
• 1 年 3 天 12 秒前是“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

Answer 1

DateJS 确实通过包含一个名为 TimeSpan 的类(在 time.js 中定义，而不是在基本 JS 文件中定义)来处理这个问题，它会为您进行计算。它不进行格式化，因此您需要这样的东西:

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，无法处理闰年之类的问题，但它应该能让您明白这一点。