mysql - 用户有很多计算机，计算机在不同的表中有很多属性，最好的加入方式是什么？

Question

我有一个用户表：

USERS:
ID   |   NAME  | 
----------------
1    |   JOHN  |
2    |   STEVE |

电脑桌：

COMPUTERS:
ID   |   USER_ID |
------------------
13   |     1     |
14   |     1     |

处理器表：

PROCESSORS:
ID   |   NAME             |
---------------------------
27   |   PROCESSOR TYPE 1 |
28   |   PROCESSOR TYPE 2 |

和一张硬盘表：

HARDDRIVES:
ID   |   NAME              |
---------------------------|
35   |   HARDDRIVE TYPE 25 |
36   |   HARDDRIVE TYPE 90 |

每台计算机都可以具有来自不同属性表（处理器、硬盘驱动器等）的许多属性，所以我有这样的交叉表，以将属性链接到计算机：

COMPUTER_PROCESSORS:
C_ID |  P_ID  |
--------------|
13   |   27   |
13   |   28   |
14   |   27   |

COMPUTER_HARDDRIVES:
C_ID |  H_ID  |
--------------|
13   |   35   |

因此，ID 为 1 的用户 JOHN 拥有计算机 13 和 14。计算机 13 具有处理器 27 和 28，计算机 13 具有硬盘驱动器 35。计算机 14 具有处理器 27，但没有硬盘驱动器。

给定用户的 id，我想检索该用户的计算机列表以及每个计算机属性。

我想出了一个查询，它给了我一些结果：

SELECT computers.id, processors.id AS p_id, processors.name AS p_name, harddrives.id AS h_id, harddrives.name AS h_name,
FROM computers

JOIN computer_processors ON (computer_processors.c_id = computers.id) 
JOIN processors ON (processors.id = computer_processors.p_id)

JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) 
JOIN harddrives ON (harddrives.id = computer_harddrives.h_id)

WHERE computers.user_id = 1

结果：

ID   | P_ID | P_NAME           | H_ID | H_NAME            |
-----------------------------------------------------------
13   |  27  | PROCESSOR TYPE 1 |  35  | HARDDRIVE TYPE 25 |
13   |  28  | PROCESSOR TYPE 2 |  35  | HARDDRIVE TYPE 25 |

但这有几个问题......

计算机 14 没有出现，因为它没有硬盘。 我可以以某种方式进行 OUTER JOIN 以确保所有计算机都显示出来，即使它们没有某些属性？
计算机 13 出现两次，两者的硬盘列表相同。当向计算机添加更多属性时（例如 3 块 ram），为该计算机返回的行数变得非常大，这使得它必须在应用程序代码中对结果进行排序。我可以以某种方式进行查询，将返回的两个行组合在一起吗？还是在第二行的 h_name 列中返回 NULL 的查询，以便返回的所有值都是唯一的？

编辑：我想返回的是这样的：

ID   | P_ID | P_NAME           | H_ID | H_NAME            |
-----------------------------------------------------------
13   |  27  | PROCESSOR TYPE 1 |  35  | HARDDRIVE TYPE 25 |
13   |  28  | PROCESSOR TYPE 2 |  35  | NULL              |
14   |  27  | PROCESSOR TYPE 1 | NULL | NULL              |

或者任何可以很容易地将它变成这样的数组的结果

[13] =>
    [P_NAME] =>
         [0] => PROCESSOR TYPE 1
         [1] => PROCESSOR TYPE 2
    [H_NAME] =>
         [0] => HARDDRIVE TYPE 25


[14] =>
    [P_NAME] =>
         [0] => PROCESSOR TYPE 1

score 4 · Accepted Answer

使用LEFT JOIN而不是INNER JOIN. 这两个连接彼此不同。（您当前正在使用INNER JOIN的）仅返回在表的任一侧都至少有一个匹配项的记录。这就是为什么不应该，因为它在 table 上没有匹配项。另一方面，返回左侧表上的所有记录，无论它在表的另一侧是否不匹配。computer 14COMPUTER_HARDDRIVESLEFT JOIN

SELECT  a.ID AS UserID,
        a.Name as UserName,
        b.ID as ComputerID,
        d.ID as ProcessorID,
        d.Name as ProcessorName,
        f.ID as HardDriveID,
        f.name as HardDriveName
FROM    users a
        INNER JOIN computers b
            ON a.ID = b.user_ID
        LEFT JOIN computer_processors c
            ON b.ID = c.C_ID
        LEFT JOIN PROCESSORS d
            ON c.p_ID = d.ID
        LEFT JOIN COMPUTER_HARDDRIVES e
            ON b.ID = e.c_ID
        LEFT JOIN HARDDRIVE f
            ON e.h_ID = f.ID
WHERE   a.ID = 1

请参阅 SQLFiddle 演示
有关连接的更多详细信息，请参阅SQL 连接的可视化表示

并且由于您想将行组合在一起，因此您可以利用GROUP_CONCAT()功能。基本上，它的作用是将列的值组合成逗号分隔值

SELECT  a.ID AS UserID,
        a.Name as UserName,
        b.ID as ComputerID,
        GROUP_CONCAT(DISTINCT d.ID) as ProcessorID,
        GROUP_CONCAT(DISTINCT d.Name) as ProcessorName,
        GROUP_CONCAT(DISTINCT f.ID) as HardDriveID,
        GROUP_CONCAT(DISTINCT f.name) as HardDriveName
FROM    users a
        INNER JOIN computers b
            ON a.ID = b.user_ID
        LEFT JOIN computer_processors c
            ON b.ID = c.C_ID
        LEFT JOIN PROCESSORS d
            ON c.p_ID = d.ID
        LEFT JOIN COMPUTER_HARDDRIVES e
            ON b.ID = e.c_ID
        LEFT JOIN HARDDRIVE f
            ON e.h_ID = f.ID
WHERE   a.ID = 1
GROUP BY UserID, UserName, ComputerID

请参阅 SQLFiddle 演示

score 3 · Accepted Answer

尝试：

SELECT computers.id, 
       group_concat(distinct processors.id) AS p_ids, 
       group_concat(distinct processors.name) AS p_names, 
       group_concat(distinct harddrives.id) AS h_ids, 
       group_concat(distinct harddrives.name) AS h_names
FROM computers
JOIN computer_processors ON (computer_processors.c_id = computers.id) 
JOIN processors ON (processors.id = computer_processors.p_id)
LEFT JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) 
LEFT JOIN harddrives ON (harddrives.id = computer_harddrives.h_id)
WHERE computers.user_id = 1
group by computers.id

score 1 · Accepted Answer

SELECT computers.id, processors.id AS p_id, processors.name AS p_name, harddrives.id AS h_id, harddrives.name AS h_name
FROM computers

LEFT JOIN computer_processors ON (computer_processors.c_id = computers.id) 
LEFT JOIN processors ON (processors.id = computer_processors.p_id)

LEFT JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) 
LEFT JOIN harddrives ON (harddrives.id = computer_harddrives.h_id)

WHERE computers.user_id = 1

score 0 · Accepted Answer

select u.ID user_id,
       u.name user_name
       p.id,
       p.name,
       h.id,
       h.name
from USERS u join COMPUTERS c
on u.id=c.USER_ID
join PROCESSORS p
on c.id=p.id
join HARDDRIVES h
on c.id=h.id
where u.id=1

mysql - 用户有很多计算机，计算机在不同的表中有很多属性，最好的加入方式是什么？

4 回答 4

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