Question

表格结构：

    users: uid, name
    friendships: uid1, uid2
    game_membership: gid, uid

基本上我正在尝试将查询作为gid的函数，并返回：

uid1, name1, uid2, name2

只返回给定uid下game_membership表中的gid。这是我到目前为止所得到的：

SELECT 
   f.uid1 AS id1, 
   u1.name AS name1, 
   f.uid2 AS id2, 
   u2.name AS name2
FROM 
   friendships AS f
WHERE 
   u1.uid in (SELECT uid FROM game_membership WHERE gid = <GID>)
   AND u2.uid in (SELECT uid FROM game_membership WHERE gid = <GID>)
JOIN users AS u1 ON f.uid1 = u1.uid
JOIN users AS u2 ON f.uid2 = u2.uid

我现在刚收到MySQL语法错误。谁能指出我正确的方向？我觉得我可能需要在uid表的game_membership上另外加一两个JOIN，每个uid一个？

Answer 1

我建议您加入Game_membership表而不是where子句：

SELECT 
   f.uid1 AS id1, 
   u1.name AS name1, 
   f.uid2 AS id2, 
   u2.name AS name2
FROM 
   friendships AS f
JOIN game_membership AS user1_gm ON user1_gm.uid = f.uid1
JOIN game_membership AS user2_gm ON user2_gm.uid = f.uid2
JOIN users AS u1 ON f.uid1 = u1.uid
JOIN users AS u2 ON f.uid2 = u2.uid
WHERE user1_gm.gid = <GID> AND user2_gm.gid = <GID>

Answer 2

将加入添加到 FROM 子句中，如下所示：

FROM friends f
    LEFT OUTER JOIN users u1 ON f.uid1 = u1.uid
    LEFT OUTER JOIN users u2 on f.uid2 = u2.uid

Answer 3

替代：

 SELECT 
   f.uid1 AS id1, 
   u1.name AS name1, 
   f.uid2 AS id2, 
   u2.name AS name2
FROM friendships AS f
INNER JOIN users AS u1 
   ON f.uid1 = u1.uid
INNER JOIN users AS u2 
   ON f.uid2 = u2.uid
INNER JOIN game_membership GMS
   ON GMS.UID=U1.UID
INNER JOIN game_membership GMS2
   ON GMS2.UID=U2.UID
WHERE 
   GMS.GID= <GID> and GMS2.GID=<GID>

现在它不是保罗的替代品：D

Answer 4

select uid1, u1.name as name1, uid2, u2.name as name2
from friendships 
    join game_membership as g1 on uid1 = g1.uid
    join game_membership as g2 on uid2 = g2.uid
    join users as u1 on uid1 = u1.uid
    join users as u2 on uid2 = u2.uid
where g1.gid = <GID> and g2.gid = <GID>

MySQL棘手的双/三连接（？）

4 个答案: