加入自己的表

Question

如何將此表加入自己以檢查兩個用戶是否都接受了好友請求？

mysql> select * from friendships; 
+---------+-----------+----------+ 
| user_id | friend_id | accepted | 
+---------+-----------+----------+ 
|  1 |   2 |  1 | 
|  2 |   1 |  0 | 
|  3 |   1 |  1 | 
|  1 |   3 |  1 | 
|  1 |   4 |  1 | 
|  4 |   1 |  0 | 
|  5 |   1 |  1 | 
|  1 |   5 |  0 | 
+---------+-----------+----------+ 
8 rows in set (0.00 sec) 

並拉出用戶對象。

我可以檢查user 1是否有任何優秀的朋友請求;

mysql> select * from friendships join users on friendships.friend_id = users.id where friendships.user_id = 1 and accepted = false; 
+---------+-----------+----------+----+------------+----------+--------------------+------------------+-----------------+ 
| user_id | friend_id | accepted | id | fullName | username | phone    | verificationCode | accountVerified | 
+---------+-----------+----------+----+------------+----------+--------------------+------------------+-----------------+ 
|  1 |   5 |  0 | 5 | Tom Tester | tom  | +16502222222222222 | 4444    |    1 | 
+---------+-----------+----------+----+------------+----------+--------------------+------------------+-----------------+ 
1 row in set (0.00 sec) 

但我怎麼讓他接受了請求（即user 1和user 3都已經接受了該請求）？

另外我有一個用戶表設置在後臺。

編輯：如果它可以幫助

CREATE TABLE friendships (
    user_id int, 
    friend_id int, 
    accepted boolean not null default false, 
    UNIQUE KEY friend (user_id, friend_id) 
); 

Answer 1

訣竅是，user_id和friend_id字段的組合作爲你們的友誼表的主鍵我的模式，這是2場，你需要加入表本身。顯然，在連接標準中，您必須交叉引用2個字段，因爲這兩個記錄中的角色是相反的。

select * 
from friendships f1 
inner join friendships f2 on f1.user_id=f2.friend_id and f1.friend_id=f2.user_id 
where f1.accepted=1 and f2.accepted=a1 and f1.user_id=... 

您也可以通過使用exists子查詢實現相同的輸出：

select * from friendships f1 
where f1.user_id=... and f1.accepted=1 
    and exists (select 1 from friendships f2 
       where f2.user_id=f1.friend_id and f1.friend_id=f1.user_id and f2.accepted=1)