基于@bob的回答，这将查询泛化为能够返回多个中位数，并按某些标准分组。

想想，例如，一个车场二手车的中位数销售价格，按年-月分组。

SELECT 
    period, 
    AVG(middle_values) AS 'median' 
FROM (
    SELECT t1.sale_price AS 'middle_values', t1.row_num, t1.period, t2.count
    FROM (
        SELECT 
            @last_period:=@period AS 'last_period',
            @period:=DATE_FORMAT(sale_date, '%Y-%m') AS 'period',
            IF (@period<>@last_period, @row:=1, @row:=@row+1) as `row_num`, 
            x.sale_price
          FROM listings AS x, (SELECT @row:=0) AS r
          WHERE 1
            -- where criteria goes here
          ORDER BY DATE_FORMAT(sale_date, '%Y%m'), x.sale_price
        ) AS t1
    LEFT JOIN (  
          SELECT COUNT(*) as 'count', DATE_FORMAT(sale_date, '%Y-%m') AS 'period'
          FROM listings x
          WHERE 1
            -- same where criteria goes here
          GROUP BY DATE_FORMAT(sale_date, '%Y%m')
        ) AS t2
        ON t1.period = t2.period
    ) AS t3
WHERE 
    row_num >= (count/2) 
    AND row_num <= ((count/2) + 1)
GROUP BY t3.period
ORDER BY t3.period;

在某些情况下，中位数的计算如下:

“中位数”是数字列表中按值排序时的“中间”值。对于偶数集，中位数是两个中间值的平均值。 我为此创建了一个简单的代码:

$midValue = 0;
$rowCount = "SELECT count(*) as count {$from} {$where}";

$even = FALSE;
$offset = 1;
$medianRow = floor($rowCount / 2);
if ($rowCount % 2 == 0 && !empty($medianRow)) {
  $even = TRUE;
  $offset++;
  $medianRow--;
}

$medianValue = "SELECT column as median 
               {$fromClause} {$whereClause} 
               ORDER BY median 
               LIMIT {$medianRow},{$offset}";

$medianValDAO = db_query($medianValue);
while ($medianValDAO->fetch()) {
  if ($even) {
    $midValue = $midValue + $medianValDAO->median;
  }
  else {
    $median = $medianValDAO->median;
  }
}
if ($even) {
  $median = $midValue / 2;
}
return $median;

返回的$中位数将是所需的结果:-)

上面的大多数解决方案只适用于表中的一个字段，您可能需要获得查询中多个字段的中位数(第50百分位数)。

我用这个:

SELECT CAST(SUBSTRING_INDEX(SUBSTRING_INDEX(
 GROUP_CONCAT(field_name ORDER BY field_name SEPARATOR ','),
  ',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) AS `Median`
FROM table_name;

你可以将上面例子中的“50”替换为任何百分位数，这是非常有效的。

只要确保你有足够的内存给GROUP_CONCAT，你可以改变它:

SET group_concat_max_len = 10485760; #10MB max length

详情:http://web.performancerasta.com/metrics-tips-calculating-95th-99th-or-any-percentile-with-single-mysql-query/

我使用下表的解决方案在MySQL:

CREATE TABLE transactions (
  transaction_id int , user_id int , merchant_name varchar(255), transaction_date date , amount int
);

INSERT INTO transactions (transaction_id, user_id, merchant_name, transaction_date, amount)  
VALUES (1, 1 ,'abc', '2015-08-17', 100),(2, 2, 'ced', '2015-2-17', 100),(3, 1, 'def', '2015-2-16', 121),
(4, 1 ,'ced', '2015-3-17', 110),(5, 1, 'ced', '2015-3-17', 150),(6, 2 ,'abc', '2015-4-17', 130), 
(7, 3 ,'ced', '2015-12-17', 10),(8, 3 ,'abc', '2015-8-17', 100),(9, 2 ,'abc', '2015-12-17', 140),(10, 1,'abc', '2015-9-17', 100),
(11, 1 ,'abc', '2015-08-17', 121),(12, 2 ,'ced', '2015-12-23', 130),(13, 1 ,'def', '2015-12-23', 13),(3, 4, 'abc', '2015-2-16', 120),(3, 4, 'def', '2015-2-16', 121),(3, 4, 'ced', '2015-2-16', 121);

计算“金额”列的中位数:

WITH Numbered AS 
(
SELECT *, COUNT(*) OVER () AS TotatRecords,
    ROW_NUMBER() OVER (ORDER BY amount) AS RowNum
FROM transactions
)
SELECT Avg(amount)
FROM Numbered
WHERE RowNum IN ( FLOOR((TotatRecords+1)/2), FLOOR((TotatRecords+2)/2) )
;

TotalRecords = 16 and Median = 120.5000

此查询将适用于两种情况，即偶数和奇数记录。

MariaDB / MySQL:

SELECT AVG(dd.val) as median_val
FROM (
SELECT d.val, @rownum:=@rownum+1 as `row_number`, @total_rows:=@rownum
  FROM data d, (SELECT @rownum:=0) r
  WHERE d.val is NOT NULL
  -- put some where clause here
  ORDER BY d.val
) as dd
WHERE dd.row_number IN ( FLOOR((@total_rows+1)/2), FLOOR((@total_rows+2)/2) );

Steve Cohen指出，在第一次传递之后，@rownum将包含总行数。这可用于确定中值，因此不需要第二次传递或连接。

此外，AVG(dd.val)和dd.row_number IN(…)用于在有偶数条记录时正确地产生中位数。推理:

SELECT FLOOR((3+1)/2),FLOOR((3+2)/2); -- when total_rows is 3, avg rows 2 and 2
SELECT FLOOR((4+1)/2),FLOOR((4+2)/2); -- when total_rows is 4, avg rows 2 and 3

最后，MariaDB 10.3.3+包含一个MEDIAN函数

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