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Question

我有兩個簡單的MySQL表：用戶&的關係。

關係表：

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

（的一部分）的用戶表：

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

我選擇使用此查詢朋友的朋友：

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 

但我也需要得到建議的朋友......（組內知道兩個或兩個以上直接朋友的人），並且我有這方面的問題。什麼是一種好方法（查詢，還是應該使用PHP？）來獲取推薦的朋友？

Answer 1

請考慮以下內容......本示例假設通過每個友誼插入兩行來確定往復。然而，爲了簡單起見，下面的例子並沒有檢查友誼是否有回報！

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'); 

要獲得所有奔的「的朋友 - 的 - 朋友」，我們可以做到這一點的列表...

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   | 
+--------------+ 

正如你所看到的，因爲丹是朋友亞當和查理（本的朋友）他的名字出現了兩次。

因此，要獲得DISTINCT朋友朋友列表，只需包含DISTINCT運算符。

同樣，獲得個人誰是陌生人本名單，但至少兩個本的朋友的朋友，我們可以做到這一點...

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   | 
+--------------+ 

可能有處理的幾種方法這個問題的相互作用方面，正如處理互惠本身的幾種方式一樣。

一種方法是用一個簡單的子查詢替換上述每個出現的friends表，

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

感謝@Strawberry，太糟糕了我無法使用您的解決方案了蝙蝠的權利......我的設計（oxwall）是完全不同的：

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

沒有DUP 1-> 2則2-> 1像

所以我的查詢：

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 

希望它可以幫助別人，將來