Question

我的第一个查询返回一周的平均值：

SELECT POSITION, PLAYER, SUM( POINTS ) / COUNT( DISTINCT YEAR, WEEK ) AS  'avg' 
FROM SCHEDULE WHERE WEEK = 1 GROUP BY PLAYER HAVING COUNT( DISTINCT YEAR, WEEK ) >2 
ORDER BY avg DESC

我的第二个查询返回整体平均值：

SELECT POSITION, PLAYER, SUM( POINTS ) / COUNT( DISTINCT YEAR, WEEK ) AS  'avg' 
FROM SCHEDULE GROUP BY PLAYER HAVING COUNT( DISTINCT YEAR, WEEK ) >2 
ORDER BY avg DESC

我希望从查询2中减去查询1并按降序显示结果。

Answer 1

以下内容如何：

SELECT m.POSITION, m.PLAYER, (m.Overallavg - m.WeeklyAvg) AS NewValue
FROM
(
  SELECT s1.POSITION, s1.PLAYER, SUM(s1.POINTS)/COUNT(DISTINCT s1.YEAR, s1.WEEK) AS Overallavg,
  (
    SELECT SUM(s2.POINTS)/COUNT(DISTINCT s2.YEAR, s2.WEEK)
    FROM SCHEDULE s2
    WHERE WEEK = 1 AND s2.PLAYER = S1.PLAYER
  ) AS WeeklyAvg
  FROM SCHEDULE s1
  GROUP BY s1.PLAYER 
  HAVING COUNT(DISTINCT s1.YEAR, s1.WEEK) > 2 
) m
ORDER BY (Overallavg - WeeklyAvg) DESC

Answer 2

SELECT 
    WA.POSITION, 
    WA.PLAYER, 
    WA.avg as 'wk_avg',
    OA.avg as 'overall_avg',
    WA.avg - OA.avg as 'diff'
FROM 
    (SELECT POSITION, PLAYER, SUM( POINTS ) / COUNT( DISTINCT YEAR, WEEK ) AS  'avg' 
    FROM SCHEDULE WHERE WEEK = 1 GROUP BY PLAYER HAVING COUNT( DISTINCT YEAR, WEEK ) >2 
    ) WA
INNER JOIN
    (SELECT POSITION, PLAYER,SUM( POINTS ) / COUNT( DISTINCT YEAR, WEEK ) AS  'avg' 
    FROM SCHEDULE GROUP BY PLAYER HAVING COUNT( DISTINCT YEAR, WEEK ) >2 
    ) OA
ON WA.POSITION = OA.POSITION and WA.PLAYER = OA.PLAYER
ORDER BY WA.avg - OA.avg DESC

Answer 3

except运算符将是一种方式，但mysql不接受它。

我使用NOT IN ...

进行了此查询

SELECT 
    S1.POSITION, S1.PLAYER, SUM( S1.POINTS ) / COUNT( DISTINCT S1.YEAR, S1.WEEK ) AS  'S1AVG' 
FROM 
    SCHEDULE S1 WHERE (S1.POSITION, S1.PLAYER, S1AVG) NOT IN 
    (
        SELECT 
            S2.POSITION, S2.PLAYER, SUM( S2.POINTS ) / COUNT( DISTINCT S2.YEAR, S2.WEEK ) AS  'S2AVG' 
        FROM 
            SCHEDULE S2 WHERE S2.WEEK = 1 GROUP BY S2.PLAYER HAVING COUNT( DISTINCT S2.YEAR, S2.WEEK ) > 2 
        ORDER BY S2AVG DESC
    ) 
GROUP BY S1.PLAYER 
HAVING COUNT( DISTINCT S1.YEAR, S1.WEEK ) > 2 
ORDER BY S1AVG DESC

Answer 4

不确定这是否适用于MySQL

SELECT @result = (QUERY1) - (QUERY2)

或者像这样

SELECT (QUERY1) - (QUERY2)

结合两个MySQL查询并减去结果

4 个答案: