MySQL查詢對

Question

如何選擇所有尚未與'alpha'配對的玩家？ （輸出='gamma'）。它應該工作，即使對錶是空的。

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

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

我一直在掙扎幾個小時。對於例如如果我這樣做

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

輸出是'beta'和'gamma'。在連接中，'beta'出現兩次（每次與'alpha'和'gamma'配對）。 WHERE條件消除了一行'beta'。我想要的是'貝塔'的所有行都被消除。我是新手，嘗試過GROUP BY，HAVING等等的各種組合，我無法讓它工作。

SQL Fiddle here.

Answer 1

我會打破這種分解成更小的比特和拼湊一起。首先獲得阿爾法球員的編號：

SELECT playerID 
FROM players 
WHERE name = 'alpha'; 

下一步是弄清楚如何排除。我會得到誰被選擇所有player2值，其中PLAYER1是「阿爾法」和所有PLAYER1值，其中player2是「阿爾法」這樣的搭配玩家1.我做這一切的球員名單：

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; 

一旦你這樣做了，剩下的唯一步驟是從球員那裏的人是1）沒有α和2）未在上面的列表中拉：這裏

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

像這樣的東西應該適用於這種情況。

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成對錶中的列更改位置，則可能必須在子查詢中使用CASE。

Answer 3

這應做到：

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

備用解決方案，不使用「NOT IN」。

方法：（所有用戶） - （所有現有的對手）

這樣，即使（所有現有的對手）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);