MYSQL query for cities with more customers

Question

i have to write a query which will return all cities with more customer than average number of customers of all cities. For each such city, return the country name, the city name, the number of customers. order the result by country name ascending.

Output should be

country_name, city_name, count

I tried to use sub query but I got an error

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

I then tried

Select country.country_name, city.city_name, count(customer.customer_name)
from country
inner join city on city.country_id = country.id
inner join customer on customer.city_id = city.id
group by country.country_name, city.city_name
having count(customer.customer_name) > 
(
  Select count(customer.customer_name) / count(city.city_name) as avg_per_city
  from city
  inner join customer on customer.city_id = city.id
) 

but this is also giving error.

your help would be great.

Answer 1

The second query looks almost correct. Since the city id is unique, you don't need to join to the city table when computing the average.

Try this:

Select country.country_name, city.city_name, count(customer.customer_name)
from country
inner join city on city.country_id = country.id
inner join customer on customer.city_id = city.id
group by country.country_name, city.city_name
having count(customer.customer_name) > 
(
  Select count(customer.customer_name) / count(distinct city_id) as avg_per_city
  from customer
)