社交网络 - 推荐的朋友

Question

我有两个简单的 MySQL 表：用户和关系。

关系表：

user_id int(10) unsigned    NO  PRI     
friend_id   int(10) unsigned    NO  PRI 

（部分）用户表：

id  int(10) unsigned    NO  PRI     
username    varchar(128)    NO

我使用这个查询选择朋友的朋友：

SELECT f2.friend_id, u.username
FROM relations f1
JOIN relations f2 ON f1.friend_id=f2.user_id
LEFT JOIN user u ON u.id = f2.friend_id
WHERE f2.friend_id NOT IN (select friend_id from relations where user_id=@user_id) AND f1.user_id= 2 AND f2.friend_id!= 2

但我还需要获得推荐的朋友...（小组中认识 2 个或更多直接朋友的人），我对此有疑问。什么是获得推荐朋友的好方法（查询，还是我应该使用 PHP？）？

Answer 1

考虑以下...这个例子假设互惠是通过在每个友谊中插入两行来建立的。但是，为简单起见，以下示例并未检查友谊是否得到回报！

 DROP TABLE IF EXISTS friends;

 CREATE TABLE friends
 (initiator VARCHAR(12) NOT NULL
 ,reciprocator VARCHAR(12) NOT NULL
 ,PRIMARY KEY (initiator,reciprocator)
 );

 INSERT INTO friends VALUES 
 ('Adam','Ed'),
 ('Ed','Adam'),
 ('Adam','Ben'),
 ('Ben','Adam'),
 ('Adam','Charlie'),
 ('Charlie','Adam'),
 ('Adam','Dan'),
 ('Dan','Adam'),
 ('Ed','Ben'),
 ('Ben','Ed'),
 ('Ben','Charlie'),
 ('Charlie','Ben'),
 ('Charlie','Dan'),
 ('Dan','Charlie'),
 ('Dan','Fred'),
      ('Fred','Dan'),
 ('Adam','Fred'),
      ('Fred','Adam');

要获得 Ben 的所有“朋友之友”的列表，我们可以这样做...

 SELECT y.reciprocator
   FROM friends x
   JOIN friends y
     ON y.initiator = x.reciprocator
    AND y.reciprocator <> x.initiator
   LEFT
   JOIN friends z
     ON z.reciprocator = y.reciprocator 
    AND z.initiator = x.initiator
  WHERE x.initiator = 'Ben'
    AND z.initiator IS NULL;
 +--------------+
 | reciprocator |
 +--------------+
 | Dan          |
 | Fred         |
 | Dan          |
 +--------------+

如您所见，因为 Dan 是 Adam 和 Charlie（Ben 的两个朋友）的朋友，所以他的名字出现了两次。

因此，要获取 DISTINCT 好友列表，只需包含 DISTINCT 运算符即可。

同样，要获取 Ben 不认识但至少是 Ben 的两个朋友的朋友的个人列表，我们可以这样做...

SELECT y.reciprocator
  FROM friends x
  LEFT
  JOIN friends y
    ON y.initiator = x.reciprocator
   AND y.reciprocator <> x.initiator
  LEFT
  JOIN friends z
    ON z.reciprocator = y.reciprocator
   AND z.initiator = x.initiator
 WHERE x.initiator = 'Ben'
   AND z.initiator IS NULL
 GROUP
    BY y.reciprocator
HAVING COUNT(*) >= 2;
 +--------------+
 | reciprocator |
 +--------------+
 | Dan          |
 +--------------+

可能有几种方法可以处理这个问题的互惠方面，就像有几种方法可以处理互惠本身一样。

一种方法是将上面出现的friends 表替换为一个简单的子查询，例如

SELECT y.reciprocator
  FROM (SELECT a.* FROM friends a JOIN friends b ON b.reciprocator = a.initiator AND b.initiator = a.reciprocator) x

  LEFT
  JOIN (SELECT a.* FROM friends a JOIN friends b ON b.reciprocator = a.initiator AND b.initiator = a.reciprocator) y
    ON y.initiator = x.reciprocator
   AND y.reciprocator <> x.initiator

  LEFT
  JOIN (SELECT a.* FROM friends a JOIN friends b ON b.reciprocator = a.initiator AND b.initiator = a.reciprocator) z
    ON z.reciprocator = y.reciprocator
   AND z.initiator = x.initiator

 WHERE x.initiator = 'Ben'
   AND z.initiator IS NULL
 GROUP
    BY y.reciprocator
HAVING COUNT(*) >= 2; 

Answer 2

谢谢@Strawberry，太糟糕了，我不能马上使用你的解决方案......我的设计（牛墙）完全不同：

+--------------+--------------+--------------+--------------+
| userId       | friendId     | status       | whatever     | 
+--------------+--------------+--------------+--------------+
| 1            | 3            | request      | Dan          |
| 3            | 6            | active       | 
| 1            | 7            | ignore
+--------------+

No dup 1->2 then 2->1 like

所以我的查询：

    SELECT
        y.friendId,
        COUNT(*) AS totalFriends
    FROM
        `ow_friends_friendship` x
    LEFT JOIN `ow_friends_friendship` y 
                ON y.userId = x.friendId         

    WHERE
        x.userId = xxx
    AND  y.friendId not in (SELECT userId FROM ow_friends_friendship WHERE friendId = xxx)
    AND  y.friendId not in (SELECT friendId  FROM ow_friends_friendship WHERE userId = xxx)

    GROUP BY
        y.friendId
    HAVING
        totalFriends >= 2
    ORDER BY
        totalFriends DESC

希望以后能帮到别人