如何select 使用三个表的差异？

Question

我需要运行一个脚本来修复我的 table company_menu 中的一些行。但是，我无法构建此查询来获取这些寄存器。我在此 link 中构建架构：http://sqlfiddle.com/#!9/5ab86b

下面我展示了预期的结果。

公司

id	name
1	company 1
2	company 2
3	company 3

menu_items

id	name
1	home
2	charts
3	users
4	projects

company_menu

id	company_id	menu_item_id
1	1	1
2	1	2
3	1	3
4	1	4
5	2	1
6	2	3

这是我预期的结果：

id	company_id	menu_item_id
1	2	2
2	2	4
3	3	1
4	3	2
5	3	3
6	3	4

CREATE TABLE companies(
  id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
  name VARCHAR(50)
 );
 
CREATE TABLE menu_items(
  id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
  name VARCHAR(50)
 );
 
CREATE TABLE company_menu(
  id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
  company_id INT,
  menu_item_id INT,
  FOREIGN KEY(company_id) REFERENCES companies(id),
  FOREIGN KEY(menu_item_id) REFERENCES menu_items(id)
);

INSERT INTO companies (name) VALUES ("Company 1"),("Company 2"),("Company 3");
INSERT INTO menu_items (name) VALUES ("home"),("charts"),("users"),("projects");
INSERT INTO company_menu (company_id, menu_item_id) VALUES (1, 1),(1, 2),(1,3),(1,4);
INSERT INTO company_menu (company_id, menu_item_id) VALUES (2, 1),(2,3);

Answer 1

我能想到的两种方式。不知道哪个效率更高。两者都是从全compannies-menu_items join开始得到所有可能的组合，然后切出现有的：

WHERE NOT EXISTS

select c.id company_id, m.id menu_item_id
from companies c
join menu_items m
where not exists (
  select * from company_menu where company_id = c.id and menu_item_id = m.id
);

LEFT JOIN + IS NULL:

select c.id company_id, m.id menu_item_id
from companies c
join menu_items m
left join company_menu cm on cm.company_id = c.id and cm.menu_item_id = m.id
where cm.id is null;

两者都可以在任何公司或 menu_item 列中排序。

http://sqlfiddle.com/#!9/5ab86b/11

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