Question

尝试创建执行以下操作的查询：

一个。按照他们拥有多少粉丝的顺序对所有用户进行排序 B.删除发出请求的人已经关注的用户
C.将提出请求的人已经看到的用户卸下

以下是我的尝试：

SELECT COUNT( f.FOLLOWING_UID_FK ) AS Records , u.uid, u.full_name, u.bio, u.thumb_img, u.college, u.HEADER_IMG 
   FROM users AS u, FOLLOWERS AS f 
WHERE u.uid <>1580
AND NOT u.uid IN ( SELECT UID_FK FROM FRIEND_FINDER_SWIPES WHERE UID_FK <>1580 ) 
AND NOT u.uid IN ( SELECT UID_FK FROM FOLLOWERS WHERE FOLLOWING_UID_FK <>1580 ) 
AND u.uid = f.FOLLOWING_UID_FK GROUP BY f.FOLLOWING_UID_FK ORDER BY records DESC LIMIT 0 , 30

奇怪的是这个查询有效，如果分解，例如这将按照大多数顺序生成所有用户的列表：

SELECT COUNT( f.FOLLOWING_UID_FK ) AS Records, u.uid, u.full_name, u.bio, u.thumb_img, u.college, u.HEADER_IMG
FROM users AS u, FOLLOWERS AS f
WHERE u.uid <>1580
AND u.uid = f.FOLLOWING_UID_FK
GROUP BY f.FOLLOWING_UID_FK
ORDER BY records DESC
LIMIT 0 , 30

这将使30位用户返回发出请求的人尚未关注且尚未查看的内容：

   SELECT  u.uid, u.full_name, u.bio, u.thumb_img, u.college, u.HEADER_IMG 
       FROM users AS u
    WHERE u.uid <>1580
    AND NOT u.uid IN ( SELECT UID_FK FROM FRIEND_FINDER_SWIPES WHERE UID_FK <>1580 ) 
    AND NOT u.uid IN ( SELECT UID_FK FROM FOLLOWERS WHERE FOLLOWING_UID_FK <>1580 ) 
   DESC LIMIT 0 , 30

但是当你结合2时，结果似乎至少可以说。为什么呢？

Answer 1

你的逻辑似乎很困惑，与你想到的目标最不相关。

首先，您要计算用户拥有的关注者数量。您可以在子查询中执行此操作。然后，您要删除与当前用户有关系的人。 NOT IN . . . <>有两个双重否定。这似乎不对。我不太清楚WHERE子句中子查询的方向是什么;我在你的问题中完全抛弃逻辑。

另外，请学会使用正确的JOIN语法

SELECT uf.NumFollowers, u.uid, u.full_name,
       u.bio, u.thumb_img, u.college, u.HEADER_IMG 
FROM (SELECT UID_FK, COUNT(*) as NumFollowers
      FROM FOLLOWERS f 
      GROUP BY UID_FK
     ) uf JOIN
     users u
     ON uf.uid = u.FOLLOWING_UID_FK
WHERE u.uid <> 1580 AND
      u.uid NOT IN (SELECT UID_FK FROM FRIEND_FINDER_SWIPES WHERE UID_FK = 1580 ) AND
      u.uid NOT IN (SELECT UID_FK FROM FOLLOWERS WHERE FOLLOWING_UID_FK = 1580 ) 
GROUP BY u.uid
ORDER BY NumFollowers DESC
LIMIT 0, 30

Answer 2

看来我回答了我自己的问题，我使用的是NOT而不是＆lt;＆gt; ANY，这就是我想出来的，它似乎正在起作用：

SELECT COUNT( f.FOLLOWING_UID_FK ) AS Records, u.uid, u.full_name, u.bio, u.thumb_img, u.college, u.HEADER_IMG
FROM users AS u, FOLLOWERS AS f
WHERE u.uid <>1580
AND u.uid <> ANY(

SELECT UID_FK
FROM FRIEND_FINDER_SWIPES
WHERE UID_FK <>1580
)
AND u.uid <> ANY(

SELECT UID_FK
FROM FOLLOWERS
WHERE FOLLOWING_UID_FK <>1580
)
AND u.uid = f.FOLLOWING_UID_FK
GROUP BY f.FOLLOWING_UID_FK
ORDER BY records DESC
LIMIT 0 , 30

返回最受关注的用户尚未关注（或已查看）

2 个答案: