MySQL查询对

Question

如何选择尚未与“阿尔法”配对的所有球员？ （输出=“伽马”）。 即使pairs表为空，它也应该工作。

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

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

我已经奋斗了几个小时。 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));

输出为“ beta”和“ gamma”。 在联接中，“ beta”出现两次（每个与“ alpha”和“ gamma”配对一次）。 而且WHERE条件消除了一行“ beta”。 我想要的是消除所有“ beta”行。 我对此并不陌生，并尝试了GROUP BY，HAVING等的各种组合。我无法使其正常工作。

SQL Fiddle在这里。

Answer 1

我将其分解成较小的部分并拼凑在一起。 首先获取alpha玩家的ID：

SELECT playerID
FROM players
WHERE name = 'alpha';

下一步将是找出如何排除。 我会得到与玩家1配对的所有玩家的列表。我通过选择所有player2值（其中player1为'alpha'）和所有player1值（其中player2为'alpha'）来做到这一点，如下所示：

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;

完成此操作后，剩下的唯一步骤是从玩家（其中1）不是Alpha，而2）不在上述列表中的玩家中退出：

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示例在这里。

Answer 2

这样的事情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'

    )

如果成对的玩家1和玩家2列的表的位置发生变化，则可能必须在子查询中使用CASE。

Answer 3

应该这样做：

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

不使用“ NOT IN”的替代解决方案。

方法：（所有用户）-（所有现有对手）

即使（所有现有的反对者）为NULL，这也有效。

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);