左多次連接同一個表

Question

假設我有一個可以由2個，3個或4個玩家玩的游戲。 我在三個表中跟蹤我的數據庫（MySQL 5.1）中的這樣一個游戲，如下所示。 我希望這些字段不言自明：

create table users (id int, login char(8));
create table games (id int, stime datetime, etime datetime);
create table users_games (uid int, gid int, score int);

[在游戲桌中跟蹤的兩次是開始和結束時間]

這是填充表格的一些虛擬數據：

insert into games values
(1, '2011-12-01 10:00:00', '2011-12-01 13:00:00'),
(2, '2011-12-02 11:00:00', '2011-12-01 14:00:00'),
(3, '2011-12-03 12:00:00', '2011-12-01 15:00:00'),
(4, '2011-12-04 13:00:00', '2011-12-01 16:00:00');

insert into users_games values
(101, 1, 10),
(102, 1, 11),
(101, 2, 12),
(103, 2, 13),
(104, 2, 14),
(102, 3, 15),
(103, 3, 16),
(104, 3, 17),
(105, 3, 18),
(102, 4, 19),
(104, 4, 20),
(105, 4, 21);

現在，我需要以下列格式生成報告：

gid     p1    p2    p3    p4  started ended
1      101   102               [g1]    [g1]
2      101   103   104         [g2]    [g2]
3      102   103   104   105   [g3]    [g3]
4      102   104   105         [g4]    [g4]

也就是說，該報告顯示了在同一行中玩游戲的所有玩家。 我還需要他們的得分和用戶表中的一些其他信息，但那是第2階段。:-)

我從這開始：

select g.id, g.stime, g.etime, ug1.uid, ug2.uid, ug3.uid, ug4.uid
from games g, users_games ug1, users_games ug2, users_games ug3, users_games ug4
where
g.id = ug1.gid and
ug1.gid = ug2.gid and
ug1.uid < ug2.uid and
ug2.gid = ug3.gid and
ug2.uid < ug3.uid and
ug3.gid = ug4.gid and
ug3.uid < ug4.uid

這給了我所有四個座位都被占用的所有游戲（即，上述虛擬數據中只有游戲ID 3）。 但這只是我需要的數據的一部分。

這是我的第二次嘗試：

select g.id, g.stime, g.etime, ug1.uid, ug2.uid,
    ifnull(ug3.uid, ''), ifnull(ug4.uid, '')
from ( games g, users_games ug1, users_games ug2 )
left join users_games ug3 on ug2.gid = ug3.gid and ug2.uid < ug3.uid
left join users_games ug4 on ug3.gid = ug4.gid and ug3.uid < ug4.uid
where
g.id = ug1.gid and
ug1.gid = ug2.gid and
ug1.uid < ug2.uid

這給了我14行以上的虛擬數據。 我試圖通過將ug1錨定到最低UID播放器的條目來消除一個錯誤源：

select g.id, g.stime, g.etime, ug1.uid, ug2.uid,
    ifnull(ug3.uid, ''), ifnull(ug4.uid, '')
from
( games g, users_games ug1, users_games ug2,
    (select gid as g, min(uid) as u from users_games group by g) as xx
)
left join users_games ug3 on ug2.gid = ug3.gid and ug2.uid < ug3.uid
left join users_games ug4 on ug3.gid = ug4.gid and ug3.uid < ug4.uid
where
g.id = xx.g and
ug1.uid = xx.u and
g.id = ug1.gid and
ug1.gid = ug2.gid and
ug1.uid < ug2.uid

現在我已經下降到9行，但我仍然有很多虛假的數據。 我可以看到問題 - 例如在游戲3中，ug1錨定到用戶102，仍然有三個玩家可以錨定ug2。 等等。 但我無法找到解決這個難題的方法 - 我怎么能最終實現一個查詢，它將以正確的順序和數字輸出4行？

在我看來，這應該是在其他情況下解決的問題。 將在這里感謝所有幫助。

Answer 1

您遇到的一個問題是您沒有將用戶描述為玩家1,2,3或4的字段。但是，您需要確保每個LEFT JOIN只加入一個玩家。

如果你向users_games添加一個“player_id”字段，那就變得微不足道......

SELECT
  *
FROM
  games
LEFT JOIN
  users_games      AS p1
    ON  p1.gid = games.id
    AND p1.player_id = 1
LEFT JOIN
  users_games      AS p2
    ON  p2.gid = games.id
    AND p2.player_id = 2
LEFT JOIN
  users_games      AS p3
    ON  p3.gid = games.id
    AND p3.player_id = 3
LEFT JOIN
  users_games      AS p4
    ON  p4.gid = games.id
    AND p4.player_id = 4

有一些替代方案可以避免所有LEFT JOIN，但是這個例子很好用，因為它是下一步的基礎......）

如果您無法添加此字段，則會變得更加復雜。 （SQL Server，Oracle等，可以使用ROW_NUMBER（）代理此player_id字段，MySQL不能。）

相反，您需要相關的子查詢來識別“下一個玩家”。

SELECT
  *
FROM
  games
LEFT JOIN
  users_games      AS p1
    ON  p1.gid = games.id
    AND p1.uid = (SELECT MIN(uid) FROM users_games WHERE gid = games.id)
LEFT JOIN
  users_games      AS p2
    ON  p2.gid = games.id
    AND p2.uid = (SELECT MIN(uid) FROM users_games WHERE gid = games.id AND uid > p1.uid)
LEFT JOIN
  users_games      AS p3
    ON  p3.gid = games.id
    AND p3.uid = (SELECT MIN(uid) FROM users_games WHERE gid = games.id AND uid > p2.uid)
LEFT JOIN
  users_games      AS p4
    ON  p4.gid = games.id
    AND p4.uid = (SELECT MIN(uid) FROM users_games WHERE gid = games.id AND uid > p3.uid)

EDIT JOIN免費版，假設存在player_id字段......

SELECT
  games.id,
  MAX(CASE WHEN users_games.player_id = 1 THEN users_games.uid END)   AS p1_id,
  MAX(CASE WHEN users_games.player_id = 2 THEN users_games.uid END)   AS p2_id,
  MAX(CASE WHEN users_games.player_id = 3 THEN users_games.uid END)   AS p3_id,
  MAX(CASE WHEN users_games.player_id = 4 THEN users_games.uid END)   AS p4_id
FROM
  games
LEFT JOIN
  users_games
    ON users_games.gid = games.id
GROUP BY
  games.id

Answer 2

SELECT games.*,
IF(min(ifnull(ug1.uid,9999999))=9999999,null,ug1.uid) AS user1,
IF(min(ifnull(ug2.uid,9999999))=9999999,null,ug2.uid) AS user2,
IF(min(ifnull(ug3.uid,9999999))=9999999,null,ug3.uid) AS user3,
IF(min(ifnull(ug4.uid,9999999))=9999999,null,ug4.uid) AS user4
FROM games
LEFT JOIN users_games AS ug1 ON ug1.gid=games.id
LEFT JOIN users_games AS ug2 ON ug2.gid=games.id AND ug2.uid>ug1.uid
LEFT JOIN users_games AS ug3 ON ug3.gid=games.id AND ug3.uid>ug2.uid
LEFT JOIN users_games AS ug4 ON ug4.gid=games.id AND ug4.uid>ug3.uid
GROUP BY games.id

ofcourse 9999999應該是最大可能的用戶ID -1。 這將針對大分組查詢交換先前答案的子查詢。

使用您的測試數據在MySQL 5.1 Ubuntu Lucid上進行測試。

Answer 3

只是.....這不簡單嗎？

SELECT g.id, GROUP_CONCAT(u.login ORDER BY u.login), g.stime, g.etime
FROM games g,
users u,
users_games ug
WHERE ug.gid=g.id
AND ug.uid=u.id
GROUP BY g.id, g.stime, g.etime

如果你想得分，只需添加一個功能，然后......

SELECT g.id, GROUP_CONCAT(
     CONCAT(u.login, '=', get_score(u.login, g.id)) ORDER BY 1
     ), g.stime, g.etime
FROM games g,
users u,
users_games ug
WHERE ug.gid=g.id
AND ug.uid=u.id
GROUP BY g.id, g.stime, g.etime