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',
);

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';
}

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。

我修改了原来的函数一点(在我看来更有用，或更符合逻辑)。

// display "X time" ago, $rcs is precision depth
function time_ago ($tm, $rcs = 0) {
  $cur_tm = time(); 
  $dif = $cur_tm - $tm;
  $pds = array('second','minute','hour','day','week','month','year','decade');
  $lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);

  for ($v = count($lngh) - 1; ($v >= 0) && (($no = $dif / $lngh[$v]) <= 1); $v--);
    if ($v < 0)
      $v = 0;
  $_tm = $cur_tm - ($dif % $lngh[$v]);

  $no = ($rcs ? floor($no) : round($no)); // if last denomination, round

  if ($no != 1)
    $pds[$v] .= 's';
  $x = $no . ' ' . $pds[$v];

  if (($rcs > 0) && ($v >= 1))
    $x .= ' ' . $this->time_ago($_tm, $rcs - 1);

  return $x;
}

我试过了，效果很好

$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-10');
$difference = $datetime1->diff($datetime2);
echo formatOutput($difference);

function formatOutput($diff){
    /* function to return the highrst defference fount */
    if(!is_object($diff)){
        return;
    }

    if($diff->y > 0){
        return $diff->y .(" year".($diff->y > 1?"s":"")." ago");
    }

    if($diff->m > 0){
        return $diff->m .(" month".($diff->m > 1?"s":"")." ago");
    }

    if($diff->d > 0){
        return $diff->d .(" day".($diff->d > 1?"s":"")." ago");
    }

    if($diff->h > 0){
        return $diff->h .(" hour".($diff->h > 1?"s":"")." ago");
    }

    if($diff->i > 0){
        return $diff->i .(" minute".($diff->i > 1?"s":"")." ago");
    }

    if($diff->s > 0){
        return $diff->s .(" second".($diff->s > 1?"s":"")." ago");
    }
}

在这里查看这个链接作为参考

谢谢!玩得开心。

function time_elapsed_string($ptime)
{
    $etime = time() - $ptime;

    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)
    {
        $d = $etime / $secs;
        if ($d >= 1)
        {
            $r = round($d);
            return $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ago';
        }
    }
}