PHP:从时间戳产生相对日期/时间 [英] PHP: producing relative date/time from timestamps

问题描述

我基本上是在尝试将 Unix 时间戳(time() 函数)转换为与过去和未来日期兼容的相对日期/时间.所以输出可能是:

I'm basically trying to convert a Unix timestamp (the time() function) to a relative date/time that's both compatible with past and future date. So outputs could be:

2 周前

1 小时 60 分钟前

1 hour and 60 minutes ago

15 分 54 秒前

10 分 15 秒后

首先我尝试编写此代码，但创建了一个无法维护的巨大功能，然后我在互联网上搜索了几个小时，但我只能找到只产生一部分时间的脚本(嗯:"1小时前"没有分钟).

First I tried to code this, but made a huge unmaintainable function, and then I searched the internet for a couple of hours, yet all I can find are scripts that produce only one part of the time (e.h: "1 hour ago" without the minutes).

你有一个已经这样做的脚本吗?

Do you have a script that already does this?

推荐答案

此函数为您提供1 小时前"或明天"之类的现在"和特定时间戳"之间的结果.

This function gives you "1 hour ago" or "Tomorrow" like results between 'now' and 'specific timestamp'.

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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