Question

所以我有3张桌子：

Member (
  memberId integer,
  name varchar(50)
)

Photo (
  memberId integer,
  photoId integer
)

Like (
  photoId integer,
  memberId integer 
)

对于每个成员，我需要得到他在所有照片中的平均喜欢如果每张照片至少有一张相似的话，我就可以这样做，但是如果我添加一张没有任何喜欢的照片，它就不会包含在结果中......

感谢您的帮助

编辑：这是我之前为memberId 201701所做的事情

SELECT AVG(RCOUNT) 
FROM 
(
    SELECT COUNT (*) AS RCOUNT 
    FROM LIKE 
    WHERE photoId IN 
    (
        SELECT photoId 
        FROM PHOTO 
        WHERE memberId = 201701
    ) 
    GROUP BY photoId 
);

Answer 1

你也可以使用finner子查询的连接（左边连接照片没有像）

SELECT AVG(T.RCOUNT) 
FROM 
(
        SELECT nvl(count( * ),0) as  RCOUNT 
        FROM PHOTO  as  a
        left  JOINN LIKE as b on a.a.photoId  = b.photoId 
        WHERE a.memberId = 201701
        group by a.photoId

) T

Answer 2

这应该可以通过表LIKE和PHOTO

之间的外部联接来实现

SELECT AVG(RCOUNT) FROM (
  SELECT COUNT (*) AS RCOUNT 
    FROM PHOTO as AS p
       LEFT OUTER JOIN LIKE as l on p.photoId = l.photoId
   WHERE p.memberId = 201701
   GROUP BY p.photoId );

Answer 3

我认为你提出了错误的问题。

如果你想得到Like表中Photo的平均值，你必须像你一样只查询Like表。
如果你想在Like表中获得Like表中的Photo的平均数，你不必查询Like表，因为Photo表包含所有PhotoId，包括Like表。

所以，之前的答案如下：

SELECT AVG(RCOUNT) FROM (
    SELECT COUNT (*) AS RCOUNT 
        FROM PHOTO as AS p
        LEFT OUTER JOIN LIKE as l on p.photoId = l.photoId
    WHERE p.memberId = 201701
GROUP BY p.photoId );

可以转换如下：

SELECT AVG(RCOUNT) FROM (
    SELECT COUNT (*) AS RCOUNT 
        FROM PHOTO as AS p
    WHERE p.memberId = 201701
    GROUP BY p.photoId );

这两个查询返回完全相同的结果，因为第一个查询使用外连接。

如何获得社交网络网站每个成员的平均评价

3 个答案: