Sql 如何获得每个社交网站成员的平均喜欢度

所以我有3张桌子：

Member (
  memberId integer,
  name varchar(50)
)

Photo (
  memberId integer,
  photoId integer
)

Like (
  photoId integer,
  memberId integer 
)

我需要为每个成员获取他所有照片中的平均喜欢度。
如果每张照片都至少有一个喜欢，我可以这样做，但是如果我添加了一张没有任何喜欢的照片，结果中不会包含它

谢谢你的帮助

编辑： 以下是我之前为memberId 201701所做的

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

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您还可以使用finner子查询的联接（对于不带like的photo，使用left联接）

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这应该可以通过表

LIKE

和

PHOTO

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 );

---

我想你问错问题了

若你们想得到Like表中照片数量的平均值，你们必须像以前一样只查询Like表

若您想获得Like表中或不在Like表中的Photo计数的平均值，则不必查询Like表，因为Photo表具有包括Like表在内的所有PhotoId

因此，前面的答案如下：

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 );

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

可以按如下方式进行转换：

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 );

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部联接。

请发布您尝试使用的查询。我在帖子中添加了它谢谢您的帮助！