我知道这里有几个答案，但这是我想到的。这只处理MySQL DATETIME值，根据我回应的原始问题。数组$a需要做一些工作。我欢迎就如何改进提出意见。电话为:

Echo time_elapsed_string('2014-11-14 09:42:28');

function time_elapsed_string($ptime)
{
    // Past time as MySQL DATETIME value
    $ptime = strtotime($ptime);

    // Current time as MySQL DATETIME value
    $csqltime = date('Y-m-d H:i:s');

    // Current time as Unix timestamp
    $ctime = strtotime($csqltime); 

    // Elapsed time
    $etime = $ctime - $ptime;

    // If no elapsed time, return 0
    if ($etime < 1){
        return '0 seconds';
    }

    $a = array( 365 * 24 * 60 * 60  =>  'year',
                 30 * 24 * 60 * 60  =>  'month',
                      24 * 60 * 60  =>  'day',
                           60 * 60  =>  'hour',
                                60  =>  'minute',
                                 1  =>  'second'
    );

    $a_plural = array( 'year'   => 'years',
                       'month'  => 'months',
                       'day'    => 'days',
                       'hour'   => 'hours',
                       'minute' => 'minutes',
                       'second' => 'seconds'
    );

    foreach ($a as $secs => $str){
        // Divide elapsed time by seconds
        $d = $etime / $secs;
        if ($d >= 1){
            // Round to the next lowest integer 
            $r = floor($d);
            // Calculate time to remove from elapsed time
            $rtime = $r * $secs;
            // Recalculate and store elapsed time for next loop
            if(($etime - $rtime)  < 0){
                $etime -= ($r - 1) * $secs;
            }
            else{
                $etime -= $rtime;
            }
            // Create string to return
            $estring = $estring . $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ';
        }
    }
    return $estring . ' ago';
}

下面是一个非常简单和非常有效的解决方案。

function timeElapsed($originalTime){

        $timeElapsed=time()-$originalTime;

        /*
          You can change the values of the following 2 variables 
          based on your opinion. For 100% accuracy, you can call
          php's cal_days_in_month() and do some additional coding
          using the values you get for each month. After all the
          coding, your final answer will be approximately equal to
          mine. That is why it is okay to simply use the average
          values below.
        */
        $averageNumbDaysPerMonth=(365.242/12);
        $averageNumbWeeksPerMonth=($averageNumbDaysPerMonth/7);

        $time1=(((($timeElapsed/60)/60)/24)/365.242);
        $time2=floor($time1);//Years
        $time3=($time1-$time2)*(365.242);
        $time4=($time3/$averageNumbDaysPerMonth);
        $time5=floor($time4);//Months
        $time6=($time4-$time5)*$averageNumbWeeksPerMonth;
        $time7=floor($time6);//Weeks
        $time8=($time6-$time7)*7;
        $time9=floor($time8);//Days
        $time10=($time8-$time9)*24;
        $time11=floor($time10);//Hours
        $time12=($time10-$time11)*60;
        $time13=floor($time12);//Minutes
        $time14=($time12-$time13)*60;
        $time15=round($time14);//Seconds

        $timeElapsed=$time2 . 'yrs ' . $time5 . 'months ' . $time7 . 
                     'weeks ' . $time9 .  'days ' . $time11 . 'hrs '
                     . $time13 . 'mins and ' . $time15 . 'secs.';

        return $timeElapsed;

}

回显时间已用（1201570814）;

样例输出:

6年4个月3周4天12小时40分36秒。

我想有一个荷兰版本，支持单复数。仅仅在结尾加一个“s”是不够的，我们用的是完全不同的词，所以我重写了这篇文章的顶部答案。

这将导致:

2年1个月2周1天1分2秒

or

1年2个月1周2天1分1秒

    public function getTimeAgo($full = false){

    $now = new \DateTime;
    $ago = new \DateTime($this->datetime());
    $diff = $now->diff($ago);

    $diff->w = floor($diff->d / 7);
    $diff->d -= $diff->w * 7;

    $string = array(
        'y' => 'jaren',
        'm' => 'maanden',
        'w' => 'weken',
        'd' => 'dagen',
        'h' => 'uren',
        'i' => 'minuten',
        's' => 'seconden',
    );
    $singleString = array(
        'y' => 'jaar',
        'm' => 'maand',
        'w' => 'week',
        'd' => 'dag',
        'h' => 'uur',
        'i' => 'minuut',
        's' => 'seconde',
    );
    // M.O. 2022-02-11 I rewrote this function to support dutch singles and plurals. Added some docs for next programmer to break his brain :)
    // For each possible notation, if corresponding value of current key is true (>1) otherwise remove its key/value from array
    // If the value from current key is 1, use value from $singleString array. Otherwise use value from $string array
    foreach ($string as $k => &$v) {
        if ($diff->$k) {
            if($diff->$k == 1){
                $v = $diff->$k . ' ' . $singleString[$k];
            } else {
                $v = $diff->$k . ' ' . $v;
            }
        } else {
            if($diff->$k == 1){
                unset($singleString[$k]);
            } else {
                unset($string[$k]);
            }
        }
    }

    // If $full = true, print all values.
    // Values have already been filtered with foreach removing keys that contain a 0 as value
    if (!$full) $string = array_slice($string, 0, 1);
    return $string ? implode(', ', $string) . '' : 'zojuist';
}

你应该先测试一下，因为我不是一个优秀的程序员:)

我不知道为什么还没有人提到碳。

https://github.com/briannesbitt/Carbon

这实际上是php dateTime的扩展(这里已经使用了)，它有:diffForHumans方法。所以你所需要做的就是:

$dt = Carbon::parse('2012-9-5 23:26:11.123789');
echo $dt->diffForHumans();

更多例子:http://carbon.nesbot.com/docs/#api-humandiff

这个解决方案的优点:

它适用于未来的日期，并将在2个月等时间内返回。 你可以使用本地化来获取其他语言，多元化也很好 如果你开始用碳来做其他事情，处理日期就会变得非常容易。

举个例子:

echo time_elapsed_string('2013-05-01 00:22:35');
echo time_elapsed_string('@1367367755'); # timestamp input
echo time_elapsed_string('2013-05-01 00:22:35', true);

输入可以是任何受支持的日期和时间格式。

输出:

4 months ago
4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago

功能:

function time_elapsed_string($datetime, $full = false) {
    $now = new DateTime;
    $ago = new DateTime($datetime);
    $diff = $now->diff($ago);

    $diff->w = floor($diff->d / 7);
    $diff->d -= $diff->w * 7;

    $string = array(
        'y' => 'year',
        'm' => 'month',
        'w' => 'week',
        'd' => 'day',
        'h' => 'hour',
        'i' => 'minute',
        's' => 'second',
    );
    foreach ($string as $k => &$v) {
        if ($diff->$k) {
            $v = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
        } else {
            unset($string[$k]);
        }
    }

    if (!$full) $string = array_slice($string, 0, 1);
    return $string ? implode(', ', $string) . ' ago' : 'just now';
}

我发现结果如下:

1年2个月0天0小时53分1秒

正因为如此，我实现了一个尊重复数的函数，删除空值，并有可能缩短输出:

function since($timestamp, $level=6) {
    global $lang;
    $date = new DateTime();
    $date->setTimestamp($timestamp);
    $date = $date->diff(new DateTime());
    // build array
    $since = array_combine(array('year', 'month', 'day', 'hour', 'minute', 'second'), explode(',', $date->format('%y,%m,%d,%h,%i,%s')));
    // remove empty date values
    $since = array_filter($since);
    // output only the first x date values
    $since = array_slice($since, 0, $level);
    // build string
    $last_key = key(array_slice($since, -1, 1, true));
    $string = '';
    foreach ($since as $key => $val) {
        // separator
        if ($string) {
            $string .= $key != $last_key ? ', ' : ' ' . $lang['and'] . ' ';
        }
        // set plural
        $key .= $val > 1 ? 's' : '';
        // add date value
        $string .= $val . ' ' . $lang[ $key ];
    }
    return $string;
}

看起来好多了:

1年2个月53分1秒

可以选择使用$level = 2来缩短它，如下所示:

1年2个月

如果你只需要它的英文版本，就删除$lang部分，或者编辑这个翻译来满足你的需要:

$lang = array(
    'second' => 'Sekunde',
    'seconds' => 'Sekunden',
    'minute' => 'Minute',
    'minutes' => 'Minuten',
    'hour' => 'Stunde',
    'hours' => 'Stunden',
    'day' => 'Tag',
    'days' => 'Tage',
    'month' => 'Monat',
    'months' => 'Monate',
    'year' => 'Jahr',
    'years' => 'Jahre',
    'and' => 'und',
);