总而言之，我有几家拥有多个网站和多个部门的公司。每个部门都有一个或多个站点。我正在重复使用所有公司的部门和网站表。我基本上拥有一个全球性的分部和网站，供每家公司使用。我的最终目标是拉动所有属于公司部门的网站。
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 |
+----+-------------+
位点
+----+-------------+
| id | name |
+----+-------------+
| 1 | site 1 |
+----+-------------+
| 2 | site 2 |
+----+-------------+
| 3 | site 3 |
+----+-------------+
司
+----+-----------------+
| id | name |
+----+-----------------+
| 1 | division 1 |
+----+-----------------+
| 2 | division 2 |
+----+-----------------+
| 3 | division 3 |
+----+-----------------+
company_divisions
+----------+--------------+
| company | division |
+----------+--------------+
| 1 | 1 |
+----------+--------------+
| 1 | 2 |
+----------+--------------+
| 2 | 1 |
+----------+--------------+
| 2 | 3 |
+----------+--------------+
company_sites
+----------+------------+
| company | site |
+----------+------------+
| 1 | 1 |
+----------+------------+
| 1 | 2 |
+----------+------------+
| 2 | 2 |
+----------+------------+
| 2 | 3 |
+----------+------------+
我原本以为我可以选择受sites和company.id约束的所有division.id，但我没有这样的运气。我尝试过子查询：
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有关，而与division无关。我似乎无法弄清楚如何让company_divisions表涉及......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
如何添加其他条件或查询，以确保用于选择company.id中的site的{{1}}与company_sites中的company.id相关company_division 1}}通过division.id？
例如，给定company.id = 2和division.id = 3我会得到site 2和site 3的结果。
总是欢迎建设性的批评。
答案 0 :(得分：2)
对于这样的树结构，我可能会删除company_divisions和company_sites表并将其设计为这样。
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
会给出
+-------------+------------+-----------+
| 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 |
+-------------+------------+-----------+
从此处开始过滤应该非常简单，只需添加WHERE条件。
答案 1 :(得分：1)
我建议这样，如果你可以修改结构并有单独的分区站点映射表。这也将允许站点属于多个站点的多个站点。
公司表：
+----+-------------+
| id | name |
+----+-------------+
| 1 | company_1 |
+----+-------------+
| 2 | company_2 |
+----+-------------+
分部表：
+----+-------------+--------------+
| id | name | company_id |
+----+-------------+ -------------+
| 1 | division 1 | 1 |
+----+-------------+--------------+
| 2 | division 2 | 2 |
+----+-------------+--------------+
| 3 | division 3 | 2 |
+----+-------------+--------------+
| 4 | division 4 | 3 |
+----+-------------+--------------+
网站表：
+----+----------------------+
| 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 |
+----+-------------+--------------+
所以你可以拥有类似的东西：
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
哪会返回
+----+-------------+-------------------+-----------------------+
| company name | division name | Url's |
+----+-------------+-------------------+------------------------
| company_1 | div 1 | http:\www.url1.com, |
| | | http:\www.url2.co |
+------------------+-------------------+-----------------------+
答案 2 :(得分：0)
您提供的类似代码，检查company_site.company = 3，这是一个在您的表中不存在的值。与company_divisions表类似。事实上，没有公司3.尝试存在的数据