PostgreSQL: get the min of a column with it's associated city - sql

Question

I have been at this for the past two hours and have tried many different ways in regards to subquery and joins. Here's the exact question "Get the name and city of customers who live in the city where the least number of products are made"
Here is a snapshot of the database tables
I know how to get the min
select min(quantity)
from products
but this returns just the min without the city attached to it so I can't search for the city in the customers table.
I have also tried group by and found it gave me 3 min's (one for each group of cities) which i believe may help me
select city,min(quantity)
from products
group by city
Putting everything together I got something that looks like
SELECT
c.name,c.city
FROM
customers c
INNER JOIN
(
SELECT
city,
MIN(quantity) AS min_quantity
FROM
products
GROUP BY
city
) AS SQ ON
SQ.city = c.city
But this returns multiple customers, which isn't correct. I assume by looking at the database the city when the lowest number of products seems to be Newark and there are no customers who reside in Newark so I assume again this query would result in 0 hits.Thank you for your time.
Example
Here is an example "Get the pids of products ordered through any agent who makes at least one order for a customer in Kyoto"
and the answer I provided is
select pid
from orders
inner join agents
on orders.aid = agents.aid
inner join customers
on customers.cid = orders.cid
where customers.city = 'Kyoto'

Answer 1

In Postgresql you have sophisticated tools, viz., windowing and CTEs.
WITH
find_least_sumq AS
(SELECT city, RANK() OVER ( PARTITION BY city ORDER BY SUM(quantity) ) AS r
FROM products)
SELECT name, city
FROM customers NATURAL JOIN find_least_sumq /* ON city */
WHERE r=1; /* rank 1 is smallest summed quantity including ties */
In Drew's answer, you are zeronig in on the cities where the smallest number of any particular item is made. I interpret the question as wanting the sum of items made in that city.

Answer 2

I guess it be something around this idea:
select customers.name, city.city, city.min
from customers
join (
select city, sum (quantity) as min
from products
group by city
--filter by the cities where the total_quantity = min_quantity
having sum (quantity) = (
--get the minimum quantity
select min(quantity) from products
)
) city on customers.city = city.city

Answer 3

This can be made so much simpler. Just sort the output by the field you want to get the minimum of.
SELECT city, quantity FROM customers ORDER BY quantity LIMIT 1;

Answer 4

I have just figured out my own answer. I guess taking a break and coming back to it was all I needed. For future readers this answer will use a subquery to help you get the min of a column and compare a different column (of that same row) to a different tables column.
This example is getting the city where the least number of products are made (quantity column) in the products table and comparing that city to the cities to the city column in the customers table, then printing the names and the city of those customers. (to help clarify, use the link in the original question to look at the structure of the database I am talking about) First step is to sum all the products to their respective cities, and then take the min of that, and then find the customers in that city.Here was my solution
with citySum as(
select city,sum(quantity) as sum
from products
group by city)
select name,city
from customers
where city
in
(select city
from citySum
where sum =(
select min(sum)
from citySum))
Here is another solution I have found today that works as well using only Sub queries
select c.name,c.city
from customers c
where c.city
in
(select city
from
(select p.city,sum(p.quantity) as lowestSum
from products p
group by p.city) summedCityQuantities
order by lowestsum asc
limit 1)