我使用了两个查询方法:

第一个得到count, min, Max和avg 第二个语句(预处理语句)使用“LIMIT @count/ 2,1”和“ORDER BY ..”子句来获得中值

它们被包装在函数defn中，因此可以从一次调用中返回所有值。

如果您的范围是静态的，并且数据不经常更改，那么预先计算/存储这些值并使用存储的值，而不是每次都从头查询，可能会更有效。

一个简单的方法来计算中位数在MySQL

set @ct := (select count(1) from station);
set @row := 0;

select avg(a.val) as median from 
(select * from  table order by val) a
where (select @row := @row + 1)
between @ct/2.0 and @ct/2.0 +1;

安装和使用本mysql统计函数:http://www.xarg.org/2012/07/statistical-functions-in-mysql/

之后，计算中值就很简单了:

SELECT median(val) FROM data;

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

ORACLE的简单解决方案:

SELECT ROUND(MEDIAN(Lat_N), 4) FROM Station;

简单的解决方案，理解MySQL:

select case MOD(count(lat_n),2) 
when 1 then (select round(S.LAT_N,4) from station S where (select count(Lat_N) from station where Lat_N < S.LAT_N ) = (select count(Lat_N) from station where Lat_N > S.LAT_N))
else (select round(AVG(S.LAT_N),4) from station S where 1 = (select count(Lat_N) from station where Lat_N < S.LAT_N ) - (select count(Lat_N) from station where Lat_N > S.LAT_N))
end from station;

解释

STATION是表名。LAT_N是具有数值的列名

假设站表中有101条记录(奇数)。这意味着如果表以asc或desc排序，则中位数是第51条记录。

In above query for every S.LAT_N of S table I am creating two tables. One for number of LAT_N values less than S.LAT_N and another for number of LAT_N values greater than S.LAT_N. Later I am comparing these two tables and if they are matched then I am selecting that S.LAT_N value. When I check for 51st records there are 50 values less than 51st record and there 50 records greater than 51st record. As you see, there are 50 records in both tables. So this is our answer. For every other record there are different number of records in two tables created for comparison. So, only 51st record meets the condition.

现在假设站表中有100条记录(偶数)。这意味着如果表以asc或desc排序，则中位数是第50条和第51条记录的平均值。

Same as odd logic I am creating two tables. One for number of LAT_N values less than S.LAT_N and another for number of LAT_N values greater than S.LAT_N. Later I am comparing these two tables and if their difference is equal to 1 then I am selecting that S.LAT_N value and find the average. When I check for 50th records there are 49 values less than 50th record and there 51 records greater than 50th record. As you see, there is difference of 1 record in both tables. So this(50th record) is our 1st record for average. Similarly, When I check for 51st records there are 50 values less than 51st record and there 49 records greater than 51st record. As you see, there is difference of 1 record in both tables. So this(51st record) is our 2nd record for average. For every other record there are different number of records in two tables created for comparison. So, only 50th and 51st records meet the condition.