mysql - 客户多于平均水平的城市。子查询

Question

我有三张 table

country_table :id int, country_name string

city_table :id int, city_name string, postal_code int, country_id int

customer_table :id int, customer_name string, city_id id, customer_address string

我正在寻找一个答案，该答案将返回客户数量超过所有城市平均客户数量的所有城市。对于每个这样的城市，返回国家名称、城市名称、客户数量。
输出应该是

country_name, city_name, count

我尝试使用子查询，但出现错误

Select country_name, city_name, count(customer_name)
from country
inner join city on city.country_id = country.id
inner join customer on customer.city_id = city.id
where customer_name > (select avg(customer_name) 
                       from customer 
                       inner join customer on customer.city_id = city.id group by id)
                       group by 1, 2

非常感谢任何帮助

Answer 1

您查询的连接逻辑看起来没问题，但是子查询需要修复。我会这样写:

Select co.country_name, ci.city_name, count(*) no_customers
from city ci
inner join country co on co.id = ci. country_id
inner join customer cu on cu.city_id = ci.id
group by co.country_id, co.country_name, ci.city_id, ci.city_name
having count(*) > (
    select count(*) / count(distinct cu1.city_id) from customer cu1
)

笔记:

无需带city子查询中的表；你可以从表customer中得到你想要的信息只要

我在 group by 中添加了主键列条款，为了处理可能的同名城市/国家(在现实生活中，确实发生过)

表别名使查询更易于读写