在 mysql 中，如何在子查询中添加另一个表连接？

Question

Table: Unit
ID NAME  VALUE
1  Kilo  1000
2  Mega  1000000
3  Giga  1000000000

Table: Storage
ID Title   Drive_value DriveUnitID Cache_value CacheUnitID Status_ID error error_unit area
1  Seagate  100         3          400         1           2         1     1          1
2  Scansoft 250         3          80          2           1         1     2          2

Table: manufac
    ID   area 
    1    US
    2    CHINA 

Table: Status
ID   Description 
1    Blah.. Blah
2    Durka Durka

期望的目标是让一个子选择连接三个表 像这样（如下），但这种语法不起作用

select 
s.title, 
t.description,
x.area,
u1.value * s.cache_value as Cache,
u2.value * s.drive_value as Drive,
u3.value * s.error_value as Error
((u4.value * s.error_value)+(u4.value * s.error_value)) as ErrorHigh
((u5.value * s.error_value)-(u5.value * s.error_value)) as ErrorHigh
from storage s
join status t on t.id = s.status_id
join manufac x on x.id = s.manufac_id 
join unit u1 on s.cache_unit_id = u1.id
join unit u2 on s.drive_unit_id = u2.id
join unit u3 on s.error_unit_id = u3.id
join unit u4 on s.error_unit_id = u4.id
join unit u5 on s.error_unit_id = u5.id

Answer 1

检查您的查询，第三个内连接没有条件。如果你要去INNER JOIN它，你不需要在你的FROM子句中包含Status！

SELECT 
s.Title,
t.Description, 
u1.VALUE * s.Drive_value AS Drive, 
u2.VALUE * s.Cache_value AS Cache
FROM Storage s
INNER JOIN Unit u1 ON u1.ID = s.DriveUnitID
INNER JOIN Unit u2 ON u2.ID = s.CacheUnitID
INNER JOIN Status t ON t.ID = s.Status_ID;

Answer 2

这些查询在语义上是相同的。但请记住，要么将表放在连接中，要么放在 from 子句中：

显式连接：

select s.title, t.description,
u1.value * s.cache_value as Cache,
u2.value * s.drive_value as Drive
from storage s
join status t on t.id = s.status_id
join unit u1 on s.cache_unit_id = u1.id
join unit u2 on s.drive_unit_id = u2.id

使用隐式连接：

select s.title, t.description,
u1.value * s.cache_value as Cache,
u2.value * s.drive_value as Drive
from storage s, status t, unit u1, unit u2
where t.id = s.status_id and s.cache_unit_id = u1.id and s.drive_unit_id = u2.id

最终用户的最终结果是相同的，并且 DBMS 也以相同的方式处理两个查询，因此使用其中一个或另一个不会增加性能。

编辑：

要求更改后，这是我认为您正在寻找的查询：

select s.title, t.description,
u1.value * s.cache_value as Cache,
u2.value * s.drive_value as Drive,
u3.value * s.error_value as Error,
u3.value * s.error_value * 2 as ErrorHigh,
u3.value * s.error_value - u3.value * s.error_value as ErrorLow,
m.area
from storage s
join status t on t.id = s.status_id
join manufac m on m.id = s.area
join unit u1 on s.cache_unit_id = u1.id
join unit u2 on s.drive_unit_id = u2.id
join unit u3 on s.error_unit_id = u3.id

Example

结果：

+----------+-------------+----------+--------------+---------+-----------+----------+-------+
|  TITLE   | DESCRIPTION |  CACHE   |    DRIVE     |  ERROR  | ERRORHIGH | ERRORLOW | AREA  |
+----------+-------------+----------+--------------+---------+-----------+----------+-------+
| Seagate  | Durka Durka |   400000 | 100000000000 |    1000 |      2000 |        0 | US    |
| Scansoft | Blah.. Blah | 80000000 | 250000000000 | 1000000 |   2000000 |        0 | CHINA |
+----------+-------------+----------+--------------+---------+-----------+----------+-------+

• 请注意，您的问题中有两列名为 ErrorHigh。
• 数学“错误”x + x = 2 * x（其中 x = u4.value * s.error_value）
• 可能是最严重的数学错误 :) x - x = 0（其中 x = u5.value * s.error_value）

所以我猜这个查询并不是你真正需要的，但应该足以引导你朝着正确的方向前进。祝你好运！