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。

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();
}

我做了这个，它工作得很好，它既适用于Unix时间戳，如1470919932，也适用于格式化时间，如16-08-11 14:53:30

function timeAgo($time_ago) {
    $time_ago =  strtotime($time_ago) ? strtotime($time_ago) : $time_ago;
    $time  = time() - $time_ago;

switch($time):
// seconds
case $time <= 60;
return 'lessthan a minute ago';
// minutes
case $time >= 60 && $time < 3600;
return (round($time/60) == 1) ? 'a minute' : round($time/60).' minutes ago';
// hours
case $time >= 3600 && $time < 86400;
return (round($time/3600) == 1) ? 'a hour ago' : round($time/3600).' hours ago';
// days
case $time >= 86400 && $time < 604800;
return (round($time/86400) == 1) ? 'a day ago' : round($time/86400).' days ago';
// weeks
case $time >= 604800 && $time < 2600640;
return (round($time/604800) == 1) ? 'a week ago' : round($time/604800).' weeks ago';
// months
case $time >= 2600640 && $time < 31207680;
return (round($time/2600640) == 1) ? 'a month ago' : round($time/2600640).' months ago';
// years
case $time >= 31207680;
return (round($time/31207680) == 1) ? 'a year ago' : round($time/31207680).' years ago' ;

endswitch;
}

?>

此函数不是为英语语言而设计的。我把这些单词翻译成英语。在用于英语之前，这需要更多的修正。

function ago($d) {
$ts = time() - strtotime(str_replace("-","/",$d));

        if($ts>315360000) $val = round($ts/31536000,0).' year';
        else if($ts>94608000) $val = round($ts/31536000,0).' years';
        else if($ts>63072000) $val = ' two years';
        else if($ts>31536000) $val = ' a year';

        else if($ts>24192000) $val = round($ts/2419200,0).' month';
        else if($ts>7257600) $val = round($ts/2419200,0).' months';
        else if($ts>4838400) $val = ' two months';
        else if($ts>2419200) $val = ' a month';


        else if($ts>6048000) $val = round($ts/604800,0).' week';
        else if($ts>1814400) $val = round($ts/604800,0).' weeks';
        else if($ts>1209600) $val = ' two weeks';
        else if($ts>604800) $val = ' a week';

        else if($ts>864000) $val = round($ts/86400,0).' day';
        else if($ts>259200) $val = round($ts/86400,0).' days';
        else if($ts>172800) $val = ' two days';
        else if($ts>86400) $val = ' a day';

        else if($ts>36000) $val = round($ts/3600,0).' year';
        else if($ts>10800) $val = round($ts/3600,0).' years';
        else if($ts>7200) $val = ' two years';
        else if($ts>3600) $val = ' a year';

        else if($ts>600) $val = round($ts/60,0).' minute';
        else if($ts>180) $val = round($ts/60,0).' minutes';
        else if($ts>120) $val = ' two minutes';
        else if($ts>60) $val = ' a minute';

        else if($ts>10) $val = round($ts,0).' second';
        else if($ts>2) $val = round($ts,0).' seconds';
        else if($ts>1) $val = ' two seconds';
        else $val = $ts.' a second';


        return $val;
    }

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';
        }
    }
}

这是我的解决方案，请检查并根据您的要求修改

function getHowLongAgo($date, $display = array('Year', 'Month', 'Day', 'Hour', 'Minute', 'Second'), $ago = '') {
        date_default_timezone_set('Australia/Sydney');
        $timestamp = strtotime($date);
        $timestamp = (int) $timestamp;
        $current_time = time();
        $diff = $current_time - $timestamp;

        //intervals in seconds
        $intervals = array(
            'year' => 31556926, 'month' => 2629744, 'week' => 604800, 'day' => 86400, 'hour' => 3600, 'minute' => 60
        );

        //now we just find the difference
        if ($diff == 0) {
            return ' Just now ';
        }

        if ($diff < 60) {
            return $diff == 1 ? $diff . ' second ago ' : $diff . ' seconds ago ';
        }

        if ($diff >= 60 && $diff < $intervals['hour']) {
            $diff = floor($diff / $intervals['minute']);
            return $diff == 1 ? $diff . ' minute ago ' : $diff . ' minutes ago ';
        }

        if ($diff >= $intervals['hour'] && $diff < $intervals['day']) {
            $diff = floor($diff / $intervals['hour']);
            return $diff == 1 ? $diff . ' hour ago ' : $diff . ' hours ago ';
        }

        if ($diff >= $intervals['day'] && $diff < $intervals['week']) {
            $diff = floor($diff / $intervals['day']);
            return $diff == 1 ? $diff . ' day ago ' : $diff . ' days ago ';
        }

        if ($diff >= $intervals['week'] && $diff < $intervals['month']) {
            $diff = floor($diff / $intervals['week']);
            return $diff == 1 ? $diff . ' week ago ' : $diff . ' weeks ago ';
        }

        if ($diff >= $intervals['month'] && $diff < $intervals['year']) {
            $diff = floor($diff / $intervals['month']);
            return $diff == 1 ? $diff . ' month ago ' : $diff . ' months ago ';
        }

        if ($diff >= $intervals['year']) {
            $diff = floor($diff / $intervals['year']);
            return $diff == 1 ? $diff . ' year ago ' : $diff . ' years ago ';
        }
    }

谢谢