PHP中的人类可读日期参考，例如“最近一个星期” [英] Human readable date reference in PHP, e.g. "last monday"

问题描述

可能重复：

人类可读的PHP当前时间敏感的日期和时间格式

使用PHP的可读日期

我正在寻找一种以会话风格生成人类可读日期引用的方法，例如：

I'm looking for a way to generate human readable date references, in conversational style, such as the following:

diff( '2012-05-15', '2012-05-21' ) == "last Tuesday"
diff( '2012-05-15', '2012-05-16' ) == "yesterday"
diff( '2012-05-15', '2012-05-17' ) == "on Tuesday"
diff( '2012-04-11', '2012-05-21' ) == "on the 11th of April"

我研究了 strtotime（），这似乎与我想要的相反。该解决方案不需要在未来的日期工作，只是过去的日期。我看到另一个问题，同样的问题要未来日期在JavaScript ，但它并没有真正解决我的问题。

I looked into strtotime(), which seems to do the inverse of what I want. The solution doesn't need to work with future dates, just past dates. I saw another question asking the same thing for future dates in JavaScript, but it didn't really solve my problem.

任何想法？

推荐答案

我认为 PHP：产生相对日期/从时间戳的时间是你正在寻找的：

I think PHP: producing relative date/time from timestamps is what you are looking for:

这是这个问题的可接受的答案的代码（从原来的问题就复制到这里链接到一个pastebin）：

This is the code from the accepted answer to this question (copied it here since the original question just links to a pastebin):

function time2str($ts)
{
    if(!ctype_digit($ts))
        $ts = strtotime($ts);

    $diff = time() - $ts;
    if($diff == 0)
        return 'now';
    elseif($diff > 0)
    {
        $day_diff = floor($diff / 86400);
        if($day_diff == 0)
        {
            if($diff < 60) return 'just now';
            if($diff < 120) return '1 minute ago';
            if($diff < 3600) return floor($diff / 60) . ' minutes ago';
            if($diff < 7200) return '1 hour ago';
            if($diff < 86400) return floor($diff / 3600) . ' hours ago';
        }
        if($day_diff == 1) return 'Yesterday';
        if($day_diff < 7) return $day_diff . ' days ago';
        if($day_diff < 31) return ceil($day_diff / 7) . ' weeks ago';
        if($day_diff < 60) return 'last month';
        return date('F Y', $ts);
    }
    else
    {
        $diff = abs($diff);
        $day_diff = floor($diff / 86400);
        if($day_diff == 0)
        {
            if($diff < 120) return 'in a minute';
            if($diff < 3600) return 'in ' . floor($diff / 60) . ' minutes';
            if($diff < 7200) return 'in an hour';
            if($diff < 86400) return 'in ' . floor($diff / 3600) . ' hours';
        }
        if($day_diff == 1) return 'Tomorrow';
        if($day_diff < 4) return date('l', $ts);
        if($day_diff < 7 + (7 - date('w'))) return 'next week';
        if(ceil($day_diff / 7) < 4) return 'in ' . ceil($day_diff / 7) . ' weeks';
        if(date('n', $ts) == date('n') + 1) return 'next month';
        return date('F Y', $ts);
    }
}

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