MySQL根据附加表的条件选择行

Question

总而言之，我有几家公司拥有多个站点和多个部门。每个部门都有一个或多个网站。我正在重复使用所有公司的部门和网站表格。我基本上拥有一个全球性的分支机构和网站，供每家公司使用。我的最终目标是将所有属于公司分部的网站都拉出来。

Company 1 
    | 
    + Division 1 
    |  | 
    |  + Site 1 
    |  + Site 2 
    | 
    + Division 2 
    | 
Company 2 
    | 
    + Division 3 
    |  | 
    |  + Site 3 
    |  + Site 2 
    | 
    + Division 1 
    | 

公司

+----+-------------+ 
| id | name  | 
+----+-------------+ 
| 1 | company 1 | 
+----+-------------+ 
| 2 | company 2 | 
+----+-------------+ 

网站

+----+-------------+ 
| id | name  | 
+----+-------------+ 
| 1 | site 1 | 
+----+-------------+ 
| 2 | site 2 | 
+----+-------------+ 
| 3 | site 3 | 
+----+-------------+ 

部门

+----+-----------------+ 
| id |  name  | 
+----+-----------------+ 
| 1 | division 1 | 
+----+-----------------+ 
| 2 | division 2 | 
+----+-----------------+ 
| 3 | division 3 | 
+----+-----------------+ 

company_divisions

+----------+--------------+ 
| company | division | 
+----------+--------------+ 
|  1 |  1  | 
+----------+--------------+ 
|  1 |  2  | 
+----------+--------------+ 
|  2 |  1  | 
+----------+--------------+ 
|  2 |  3  | 
+----------+--------------+ 

个

company_sites

+----------+------------+ 
| company | site  | 
+----------+------------+ 
|  1 |  1  | 
+----------+------------+ 
|  1 |  2  | 
+----------+------------+ 
|  2 |  2  | 
+----------+------------+ 
|  2 |  3  | 
+----------+------------+ 

我本来以为我可以选择全部由company.id和division.id制约sites的，但我已经没有这样的运气。我试图子查询：

select * 
from sites 
where id IN (select site from company_sites where company = 3) 

，并加入：

select s.* 
from sites s 
inner join company_sites cs on s.id = cs.site 
where cs.company = 3 

但这些结果仅与company_site而不是division。我似乎无法弄清楚如何获得company_divisions表参与..这样的事情：

select s.* 
from sites s 
inner join company_sites cs on s.id = cs.site 
inner join company_divisions cd on divisions.id = cd.division 
where cs.company = 2 AND cd.division = 3 

我如何添加，以确保一个附加条件或查询，同样的company.id这是用来选择site在company_sites涉及company.id在company_division通过division.id？

例如给出company.id = 2和division.id = 3我期望site 2和site 3的结果。

建设性的批评总是值得欢迎的。

Answer 1

对于这样的树结构，我可能会删除company_divisions和company_sites表并设计它。

company 
+----+-------------+ 
| id | name  | 
+----+-------------+ 
| 1 | company 1 | 
+----+-------------+ 
| 2 | company 2 | 
+----+-------------+ 
name should be unique, id is the primary key 

divisions 
+----+-----------------+-------------+ 
| id |  name  | company id | 
+----+-----------------+-------------+ 
| 1 | division 1 |  1  | 
+----+-----------------+-------------+ 
| 2 | division 2 |  1  | 
+----+-----------------+-------------+ 
| 3 | division 3 |  2  | 
+----+-----------------+-------------+ 
| 4 | division 1 |  2  | 
+----+-----------------+-------------+ 
id is the primary key, company id is foreign key referenced to company.id. 

sites 
+----+-------------+-------------+ 
| id | name  | division id | 
+----+-------------+-------------+ 
| 1 | site 1 |  1  | 
+----+-------------+-------------+ 
| 2 | site 2 |  1  | 
+----+-------------+-------------+ 
| 3 | site 3 |  3  | 
+----+-------------+-------------+ 
| 4 | site 2 |  3  | 
+----+-------------+-------------+ 
id is the primary key, division id is foreign key referenced to divisions.id. 

使用查询

SELECT 
    sites.name as `site`, 
    divisions.name as `division`, 
    company.name as `company` 
FROM sites 
LEFT JOIN divisions ON sites.`division id` = divisions.id 
LEFT JOIN company ON divisions.`company id` = company.id 

会从这里起给

+-------------+------------+-----------+ 
| site  | division | company | 
+-------------+------------+-----------+ 
| site 1 | division 1 | company 1 | 
+-------------+------------+-----------+ 
| site 2 | division 1 | company 1 | 
+-------------+------------+-----------+ 
| site 3 | division 3 | company 2 | 
+-------------+------------+-----------+ 
| site 2 | division 3 | company 2 | 
+-------------+------------+-----------+ 

过滤应该是很容易的，只需要添加WHERE条件落后。

Answer 2

您在这样的事情中给出的代码会检查company_site.company = 3，这是表中不存在的值。与company_divisions表类似。事实上，没有公司3.尝试存在的数据

Answer 3

我会建议这个，如果你可以修改结构并有单独的分割网站映射表。这也可以让一个网站属于多个分部到多个网站。

Company表：

+----+-------------+ 
| id | name  | 
+----+-------------+ 
| 1 | company_1 | 
+----+-------------+ 
| 2 | company_2 | 
+----+-------------+ 

司表：

+----+-------------+--------------+ 
| id | name  | company_id | 
+----+-------------+ -------------+ 
| 1 | division 1 |  1  | 
+----+-------------+--------------+ 
| 2 | division 2 |  2  | 
+----+-------------+--------------+ 
| 3 | division 3 |  2  | 
+----+-------------+--------------+ 
| 4 | division 4 |  3  | 
+----+-------------+--------------+ 

站点表：

+----+----------------------+ 
| id | url    | 
+----+----------------------+ 
| 1 | http:\www.url1.com | 
+----+----------------------+ 
| 2 | http:\www.url2.com | 
+----+----------------------+ 
| 3 | http:\www.url3.com | 
+----+----------------------+ 
| 4 | http:\www.url4.com | 
+----+----------------------+ 

division_site表：

+----+-------------+--------------+ 
| id | div_id | site_id  | 
+----+-------------+ -------------+ 
| 1 |  1  |  1  | 
+----+-------------+--------------+ 
| 2 |  1  |  2  | 
+----+-------------+--------------+ 
| 3 |  2  |  1  | 
+----+-------------+--------------+ 
| 4 |  2  |  3  | 
+----+-------------+--------------+ 
| 5 |  2  |  4  | 
+----+-------------+--------------+ 
| 6 |  3  |  1  | 
+----+-------------+--------------+ 
| 7 |  4  |  2  | 
+----+-------------+--------------+ 

所以，你可以有这样的：

select company.name as company_name ,division.name as division_name ,GROUP_CONCAT(sites.url) as "Site URL's" from company inner join division on division.company_id = company.id left JOIN div_sites on div_sites.div_id = division.id inner join sites on sites.id = div_sites.site_id where company.id = 1 and division.id =1 GROUP by division.id 

这将返回

+----+-------------+-------------------+-----------------------+ 
| company name  | division name | Url's    | 
+----+-------------+-------------------+------------------------ 
| company_1  | div 1   | http:\www.url1.com, | 
|     |     | http:\www.url2.co  | 
+------------------+-------------------+-----------------------+