如果你正在使用PostgreSQL，那么它将为你做的工作:

const DT_SQL = <<<SQL
WITH lapse AS (SELECT (?::timestamp(0) - now()::timestamp(0))::text t)
SELECT CASE
  WHEN (select t from lapse) ~ '^\s*-' THEN replace((select t from lapse), '-', '') ||' ago' 
  ELSE (select t from lapse) END;
SQL;

function timeSpanText($ts, $conn)
// $ts: date-time string, $conn: PostgreSQL PDO connection
{
 return $conn -> prepare(DT_SQL) -> execute([ts]) -> fetchColumn();
}

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;
}

这其实是我找到的一个更好的解决方案。使用jQuery，但它的工作完美。此外，它会自动刷新类似于SO和Facebook的方式，所以你不需要刷新页面就能看到更新。

这个插件将读取<time>标签中的datetime attr，并为您填写它。

e.g. "4 minutes ago" or "about 1 day ago

http://timeago.yarp.com/

再加上另一个选择……

虽然我更喜欢在这里发布的DateTime方法，但我不喜欢它显示0年等事实。

/* 
 * Returns a string stating how long ago this happened
 */

private function timeElapsedString($ptime){
    $diff = time() - $ptime;
    $calc_times = array();
    $timeleft   = array();

    // Prepare array, depending on the output we want to get.
    $calc_times[] = array('Year',   'Years',   31557600);
    $calc_times[] = array('Month',  'Months',  2592000);
    $calc_times[] = array('Day',    'Days',    86400);
    $calc_times[] = array('Hour',   'Hours',   3600);
    $calc_times[] = array('Minute', 'Minutes', 60);
    $calc_times[] = array('Second', 'Seconds', 1);

    foreach ($calc_times AS $timedata){
        list($time_sing, $time_plur, $offset) = $timedata;

        if ($diff >= $offset){
            $left = floor($diff / $offset);
            $diff -= ($left * $offset);
            $timeleft[] = "{$left} " . ($left == 1 ? $time_sing : $time_plur);
        }
    }

    return $timeleft ? (time() > $ptime ? null : '-') . implode(' ', $timeleft) : 0;
}

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。

这里的许多解决方案没有考虑舍入。例如:

事件发生在两天前的下午3点。如果您在下午2点查看，它会显示在一天前。如果你在下午4点查看，它会显示两天前。

如果你使用unix时间，这有助于:

// how long since event has passed in seconds
$secs = time() - $time_ago;

// how many seconds in a day
$sec_per_day = 60*60*24;

// days elapsed
$days_elapsed = floor($secs / $sec_per_day);

// how many seconds passed today
$today_seconds = date('G')*3600 + date('i') * 60 + date('s');

// how many seconds passed in the final day calculation
$remain_seconds = $secs % $sec_per_day;

if($today_seconds < $remain_seconds)
{
    $days_elapsed++;
}

echo 'The event was '.$days_ago.' days ago.';

如果你担心闰秒和夏时制，这并不完美。