你也可以选择在存储过程中这样做:

DROP PROCEDURE IF EXISTS median;
DELIMITER //
CREATE PROCEDURE median (table_name VARCHAR(255), column_name VARCHAR(255), where_clause VARCHAR(255))
BEGIN
  -- Set default parameters
  IF where_clause IS NULL OR where_clause = '' THEN
    SET where_clause = 1;
  END IF;

  -- Prepare statement
  SET @sql = CONCAT(
    "SELECT AVG(middle_values) AS 'median' FROM (
      SELECT t1.", column_name, " AS 'middle_values' FROM
        (
          SELECT @row:=@row+1 as `row`, x.", column_name, "
          FROM ", table_name," AS x, (SELECT @row:=0) AS r
          WHERE ", where_clause, " ORDER BY x.", column_name, "
        ) AS t1,
        (
          SELECT COUNT(*) as 'count'
          FROM ", table_name, " x
          WHERE ", where_clause, "
        ) AS t2
        -- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
        WHERE t1.row >= t2.count/2
          AND t1.row <= ((t2.count/2)+1)) AS t3
    ");

  -- Execute statement
  PREPARE stmt FROM @sql;
  EXECUTE stmt;
END//
DELIMITER ;


-- Sample usage:
-- median(table_name, column_name, where_condition);
CALL median('products', 'price', NULL);

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

这种方法似乎包括偶数和奇数计数，没有子查询。

SELECT AVG(t1.x)
FROM table t1, table t2
GROUP BY t1.x
HAVING SUM(SIGN(t1.x - t2.x)) = 0

通常，我们不仅需要为整个表计算Median，还需要为与ID相关的聚合计算Median。换句话说，计算表中每个ID的中位数，其中每个ID有许多记录。(良好的性能和工作在许多SQL +修复偶数和赔率的问题，更多关于不同的中值方法的性能https://sqlperformance.com/2012/08/t-sql-queries/median)

SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val, 
  COUNT(*) OVER (PARTITION BY our_id) AS cnt,
  ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rn
  FROM our_table
) AS x
WHERE rn IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;

希望能有所帮助

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.

你可以使用这里的用户定义函数。