MySQL从8.0版本开始支持窗口函数，您可以使用ROW_NUMBER或DENSE_RANK(不要使用RANK，因为它将相同的RANK分配给相同的值，就像在体育排名):

SELECT AVG(t1.val) AS median_val
  FROM (SELECT val, 
               ROW_NUMBER() OVER(ORDER BY val) AS rownum
          FROM data) t1,
       (SELECT COUNT(*) AS num_records FROM data) t2
 WHERE t1.row_num IN
       (FLOOR((t2.num_records + 1) / 2), 
        FLOOR((t2.num_records + 2) / 2));

来自: http://mdb-blog.blogspot.com/2015/06/mysql-find-median-nth-element-without.html

我建议另一种方法，不用连接， 但是处理字符串

我没有用大数据表来检验， 但是中小型的桌子就可以了。

这里的好处是，它也可以通过GROUPING工作，所以它可以返回几个项目的中值。

下面是测试表的测试代码:

DROP TABLE test.test_median
CREATE TABLE test.test_median AS
SELECT 'book' AS grp, 4 AS val UNION ALL
SELECT 'book', 7 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 9 UNION ALL
SELECT 'book', 8 UNION ALL
SELECT 'book', 3 UNION ALL

SELECT 'note', 11 UNION ALL

SELECT 'bike', 22 UNION ALL
SELECT 'bike', 26 

求每组中位数的代码:

SELECT grp,
         SUBSTRING_INDEX( SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val), ',', COUNT(*)/2 ), ',', -1) as the_median,
         GROUP_CONCAT(val ORDER BY val) as all_vals_for_debug
FROM test.test_median
GROUP BY grp

输出:

grp | the_median| all_vals_for_debug
bike| 22        | 22,26
book| 4         | 2,2,3,4,7,8,9
note| 11        | 11

归档完美中位数的单个查询:

SELECT 
COUNT(*) as total_rows, 
IF(count(*)%2 = 1, CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL), ROUND((CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) + CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL)) / 2)) as median, 
AVG(val) as average 
FROM 
data

我有一个包含大约10亿行的数据库，我们需要它来确定集合中的年龄中位数。对十亿行进行排序是困难的，但如果你将可以找到的不同值(年龄范围从0到100)聚合在一起，你可以对这个列表进行排序，并使用一些算术魔术来找到你想要的任何百分位数，如下所示:

with rawData(count_value) as
(
    select p.YEAR_OF_BIRTH
        from dbo.PERSON p
),
overallStats (avg_value, stdev_value, min_value, max_value, total) as
(
  select avg(1.0 * count_value) as avg_value,
    stdev(count_value) as stdev_value,
    min(count_value) as min_value,
    max(count_value) as max_value,
    count(*) as total
  from rawData
),
aggData (count_value, total, accumulated) as
(
  select count_value, 
    count(*) as total, 
        SUM(count(*)) OVER (ORDER BY count_value ROWS UNBOUNDED PRECEDING) as accumulated
  FROM rawData
  group by count_value
)
select o.total as count_value,
  o.min_value,
    o.max_value,
    o.avg_value,
    o.stdev_value,
    MIN(case when d.accumulated >= .50 * o.total then count_value else o.max_value end) as median_value,
    MIN(case when d.accumulated >= .10 * o.total then count_value else o.max_value end) as p10_value,
    MIN(case when d.accumulated >= .25 * o.total then count_value else o.max_value end) as p25_value,
    MIN(case when d.accumulated >= .75 * o.total then count_value else o.max_value end) as p75_value,
    MIN(case when d.accumulated >= .90 * o.total then count_value else o.max_value end) as p90_value
from aggData d
cross apply overallStats o
GROUP BY o.total, o.min_value, o.max_value, o.avg_value, o.stdev_value
;

这个查询取决于你的db支持窗口函数(包括ROWS UNBOUNDED precede)，但如果你没有，这是一个简单的事情，将aggData CTE与自身连接，并将所有先前的总数聚合到' cumulative '列，用于确定哪个值包含指定的预分词。上面的示例计算p10、p25、p50(中位数)、p75和p90。

屁股的

我刚刚在网上的评论中找到了另一个答案:

对于几乎所有SQL中的中位数: SELECT x.val from data x, data y GROUP BY x.val 总和(符号(1-SIGN (y.val-x.val))) = (COUNT (*) + 1) / 2

确保列有良好的索引，并且索引用于筛选和排序。与解释计划核对。

select count(*) from table --find the number of rows

计算“中值”行号。可能使用:median_row = floor(count / 2)。

然后把它从列表中挑出来:

select val from table order by val asc limit median_row,1

这将返回您想要的值的一行。

因为我只需要一个中位数和百分位数的解决方案，我根据这个线程中的发现做了一个简单而相当灵活的函数。我知道，如果我发现“现成的”功能很容易包含在我的项目中，我自己会很高兴，所以我决定快速分享:

function mysql_percentile($table, $column, $where, $percentile = 0.5) {

    $sql = "
            SELECT `t1`.`".$column."` as `percentile` FROM (
            SELECT @rownum:=@rownum+1 as `row_number`, `d`.`".$column."`
              FROM `".$table."` `d`,  (SELECT @rownum:=0) `r`
              ".$where."
              ORDER BY `d`.`".$column."`
            ) as `t1`, 
            (
              SELECT count(*) as `total_rows`
              FROM `".$table."` `d`
              ".$where."
            ) as `t2`
            WHERE 1
            AND `t1`.`row_number`=floor(`total_rows` * ".$percentile.")+1;
        ";

    $result = sql($sql, 1);

    if (!empty($result)) {
        return $result['percentile'];       
    } else {
        return 0;
    }

}

使用非常简单，例子来自我目前的项目:

...
$table = DBPRE."zip_".$slug;
$column = 'seconds';
$where = "WHERE `reached` = '1' AND `time` >= '".$start_time."'";

    $reaching['median'] = mysql_percentile($table, $column, $where, 0.5);
    $reaching['percentile25'] = mysql_percentile($table, $column, $where, 0.25);
    $reaching['percentile75'] = mysql_percentile($table, $column, $where, 0.75);
...