这些方法从同一个表中选择两次。如果源数据来自一个昂贵的查询，这是一种避免运行两次的方法:

select KEY_FIELD, AVG(VALUE_FIELD) MEDIAN_VALUE
from (
    select KEY_FIELD, VALUE_FIELD, RANKF
    , @rownumr := IF(@prevrowidr=KEY_FIELD,@rownumr+1,1) RANKR
    , @prevrowidr := KEY_FIELD
    FROM (
        SELECT KEY_FIELD, VALUE_FIELD, RANKF
        FROM (
            SELECT KEY_FIELD, VALUE_FIELD 
            , @rownumf := IF(@prevrowidf=KEY_FIELD,@rownumf+1,1) RANKF
            , @prevrowidf := KEY_FIELD     
            FROM (
                SELECT KEY_FIELD, VALUE_FIELD 
                FROM (
                    -- some expensive query
                )   B
                ORDER BY  KEY_FIELD, VALUE_FIELD
            ) C
            , (SELECT @rownumf := 1) t_rownum
            , (SELECT @prevrowidf := '*') t_previd
        ) D
        ORDER BY  KEY_FIELD, RANKF DESC
    ) E
    , (SELECT @rownumr := 1) t_rownum
    , (SELECT @prevrowidr := '*') t_previd
) F
WHERE RANKF-RANKR BETWEEN -1 and 1
GROUP BY KEY_FIELD

下面的查询对于奇数行和偶数行都非常有效。在子查询中，我们正在寻找前后行数相同的值。对于奇数行的情况，having子句的值将为0(前后相同的行数将抵消符号)。

类似地，对于偶数行，having子句对于两行(中间的两行)的计算结果为1，因为它们(总的来说)前后的行数相同。

在外层查询中，我们将平均出单个值(奇数行)或(偶数行2个值)。

select avg(val) as median
from
(
    select d1.val
    from data d1 cross join data d2
    group by d1.val
    having abs(sum(sign(d1.val-d2.val))) in (0,1)
) sub

注意:如果你的表有重复的值，上面的having子句应该更改为下面的条件。在这种情况下，可能有一些值超出了原来的可能性(0,1)下面的条件将使这个条件动态，并在重复的情况下工作。

having sum(case when d1.val=d2.val then 1 else 0 end)>=
abs(sum(sign(d1.val-d2.val)))

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。

如果这是MySQL，现在有窗口函数，你可以这样做(假设你想四舍五入到最接近的整数-否则只需将round替换为CEIL或FLOOR或其他什么)。下面的解决方案适用于表，无论表的行数是偶数还是奇数:

WITH CTE AS (
    SELECT val,
            ROW_NUMBER() OVER (ORDER BY val ASC) AS rn,
            COUNT(*) OVER () AS total_count
    FROM data
)
SELECT ROUND(AVG(val)) AS median
FROM CTE
WHERE
    rn BETWEEN
    total_count / 2.0 AND
    total_count / 2.0 + 1;

I think some of the more recent answers on this thread were already getting at this approach, but it also seemed like people were overthinking it, so consider this an improved version. Regardless of SQL flavor, there is no reason anyone should be writing a huge paragraph of code with multiple subqueries just to get the median in 2021. However, please note that the above query only works if you're asked to find the median for a continuous series. Of course, regardless of row number, sometimes people do make a distinction between what is referred to as the Discrete Median and what is referred to as the Interpolated Median for a continuous series.

如果你被要求为一个离散级数找到中位数，而表的行数是偶数，那么上面的解决方案就不适合你，你应该恢复使用其他解决方案之一，比如TheJacobTaylor的。

下面的第二个解决方案是对TheJacobTaylor的稍微修改的版本，其中我显式地声明了CROSS JOIN。这个方法也适用于行数为奇数的表，不管你是被要求求连续序列的中位数还是离散序列的中位数，但我特别会在被要求求离散序列的中位数时使用这个方法。否则，使用第一种解决方案。这样，您就永远不必考虑数据是包含“偶数”还是“奇数”个数的数据点。

SELECT x.val AS median
FROM data x
CROSS JOIN data y
GROUP BY x.val
HAVING SUM(SIGN(1 - SIGN(y.val - x.val))) = (COUNT(*) + 1) / 2;

最后，你可以在PostgreSQL中使用内置函数轻松做到这一点。这里有一个很好的解释，以及关于离散中位数和插值中位数的有效总结。

https://leafo.net/guides/postgresql-calculating-percentile.html#calculating-the-median

create table med(id integer);
insert into med(id) values(1);
insert into med(id) values(2);
insert into med(id) values(3);
insert into med(id) values(4);
insert into med(id) values(5);
insert into med(id) values(6);

select (MIN(count)+MAX(count))/2 from 
(select case when (select count(*) from 
med A where A.id<B.id)=(select count(*)/2 from med) OR 
(select count(*) from med A where A.id>B.id)=(select count(*)/2 
from med) then cast(B.id as float)end as count from med B) C;

 ?column? 
----------
  3.5
(1 row)

OR

select cast(avg(id) as float) from 
(select t1.id from med t1 JOIN med t2 on t1.id!= t2.id 
group by t1.id having ABS(SUM(SIGN(t1.id-t2.id)))=1) A;

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);
...