我想列出表格中的所有公司以及每个特定公司的管理员数量。
表格:
companies
id
name
...
users
id
company_id
...
groups ('id' = 1 has 'name' = admin)
id
name
users_groups
id
user_id
group_id
列出所有'公司'我写这个:
SELECT companies.name
FROM companies
获取' admin'在一个特定的公司'(具有给定的ID)我写这个
SELECT COUNT (users.id)
FROM users, companies, users_groups WHERE
users.company_id = companies.id AND
users_groups.user_id = users.id AND
users_groups.group_id = 1
那么如何合并这两个问题呢?这失败了:
SELECT
companies.name,
(
SELECT COUNT (users.id)
FROM users, companies, users_groups WHERE
users.company_id = companies.id AND
users_groups.user_id = users.id AND
users_groups.group_id = 1
)
as admins_in_company
FROM users, companies, users_groups
答案 0 :(得分:1)
使用显式连接语法和计数(distinct ...):
select c.name, count(distinct u.id)
from companies c
inner join users u
on u.company_id = c.id
inner join users_groups ug
on ug.user_id = u.id
where ug.group_id = 1
group by c.name
适用于所有公司:
select c.name, count(distinct u.id)
from companies c
left join users u
on u.company_id = c.id
left join users_groups ug
on ug.user_id = u.id
and ug.group_id = 1
group by c.name
答案 1 :(得分:0)
SELECT c.name, COUNT(DISTINCT u.id) AS num_admins
FROM groups AS g
JOIN users_groups AS ug ON ug.group_id = g.id
JOIN users AS u ON u.id = ug.user_id
JOIN companies AS c ON c.id = u.company_id
WHERE g.group_id = 1
AND g.name = 'admin'
GROUP BY u.company_id;
目前还不清楚您是需要COUNT(DISTINCT u.id)
还是COUNT(*)
。
我按照他们将要查看的顺序列出了4个表。 (这不是一项要求,但可以让用户更轻松地阅读和思考。)首先是groups
,其中包含所有'过滤(
WHERE ). Then it moves through the other tables all the way to getting the
公司.name . Then the
GROUP BY and its
COUNT(DISTINCT ...)`已应用。
许多提示:许多架构(users_groups)设计:http://mysql.rjweb.org/doc.php/index_cookbook_mysql#many_to_many_mapping_table
groups
- group_id and
group.name are in a 1:1 relationship, yes? If you know that it is
group_id = 1 , you can get rid of the table
个群组completely from the query. If not, then be sure to have
INDEX(姓名)`。