MySQL查询对

Question

如何选择所有尚未与'alpha'配对的玩家？ （输出='gamma'）。它应该工作，即使对表是空的。

table: players 
+----------+-------+ 
| playerID | name | 
+----------+-------+ 
|  1 | alpha | 
|  2 | beta | 
|  3 | gamma | 
+----------+-------+ 

table: pairs 
+---------+---------+ 
| player1 | player2 | 
+---------+---------+ 
|  2 |  3 | 
|  1 |  2 | 
+---------+---------+ 

我一直在挣扎几个小时。对于例如如果我这样做

SELECT p.*, r.* 
FROM players p 
JOIN pairs r 
    ON (player1 = playerID) OR (player2 = playerID) 
WHERE 
    ((r.player1 != 1) AND (r.player2 != 1)); 

输出是'beta'和'gamma'。在连接中，'beta'出现两次（每次与'alpha'和'gamma'配对）。 WHERE条件消除了一行'beta'。我想要的是'贝塔'的所有行都被消除。我是新手，尝试过GROUP BY，HAVING等等的各种组合，我无法让它工作。

SQL Fiddle here.

Answer 1

我会打破这种分解成更小的比特和拼凑一起。首先获得阿尔法球员的编号：

SELECT playerID 
FROM players 
WHERE name = 'alpha'; 

下一步是弄清楚如何排除。我会得到谁被选择所有player2值，其中PLAYER1是“阿尔法”和所有PLAYER1值，其中player2是“阿尔法”这样的搭配玩家1.我做这一切的球员名单：

SELECT p.player2 
FROM pairs p 
JOIN(
    SELECT playerID 
    FROM players 
    WHERE name = 'alpha') b ON b.playerID = p.player1 
UNION 
SELECT p.player1 
FROM pairs p 
JOIN(
    SELECT playerID 
    FROM players 
    WHERE name = 'alpha') b ON b.playerID = p.player2; 

一旦你这样做了，剩下的唯一步骤是从球员那里的人是1）没有α和2）未在上面的列表中拉：这里

SELECT * 
FROM players 
WHERE playerID NOT IN(
    SELECT playerID 
    FROM players 
    WHERE name = 'alpha') 
AND playerID NOT IN(
    SELECT p.player2 
    FROM pairs p 
    JOIN(
    SELECT playerID 
    FROM players 
    WHERE name = 'alpha') b ON b.playerID = p.player1 
    UNION 
    SELECT p.player1 
    FROM pairs p 
    JOIN(
    SELECT playerID 
    FROM players 
    WHERE name = 'alpha') b ON b.playerID = p.player2); 

SQL Fiddle例子。

Answer 2

像这样的东西应该适用于这种情况。

SELECT 
    pl.* 
FROM 
    players as pl 
WHERE 
    pl.playerID NOT IN 
    (
     SELECT 
      p.player2 
     FROM 
      pairs AS p 
     INNER JOIN 
      players plr ON plr.playerID = p.player1 
     WHERE 
      plr.name='alpha' 

    ) 

如果玩家1和玩家2成对表中的列更改位置，则可能必须在子查询中使用CASE。

Answer 3

这应做到：

drop table pair; 
drop table player; 

create table player (
    id int, 
    name varchar(32) 
); 

create table pair (
    id1 int, 
    id2 int 
); 

insert into player values (1, 'Andy'); 
insert into player values (2, 'Bob'); 
insert into player values (3, 'Carl'); 
insert into player values (4, 'Dave'); 

insert into pair values (2, 3); 
insert into pair values (1, 2); 
insert into pair values (3, 1); 

select * from player where id not in (
select 
    if(pair.id1 = player.id, pair.id2, pair.id1) as other_player 
from 
    pair 
    join player on pair.id1 = player.id or pair.id2 = player.id 
where 
    player.name = 'Andy' 
) 
; 

Answer 4

备用解决方案，不使用“NOT IN”。

方法：（所有用户） - （所有现有的对手）

这样，即使（所有现有的对手）IS NULL。

SELECT p.* FROM players p 
LEFT JOIN (
    SELECT CASE 
      WHEN (player1 = 1) THEN player2 
      WHEN (player2 = 1) THEN player1 
     END AS opponentID 
    FROM pairs) existingOpponents 
ON (p.playerID = existingOpponents.opponentID) 
WHERE (existingOpponents.opponentID IS NULL) AND 
    (p.playerID != 1);