SQL How to select customers with highest transaction amount by state - sql

Question

I am trying to write a SQL query that returns the name and purchase amount of the five customers in each state who have spent the most money.
Table schemas
customers
|_state
|_customer_id
|_customer_name
transactions
|_customer_id
|_transact_amt
Attempts look something like this
SELECT state, Sum(transact_amt) AS HighestSum
FROM (
SELECT name, transactions.transact_amt, SUM(transactions.transact_amt) AS HighestSum
FROM customers
INNER JOIN customers ON transactions.customer_id = customers.customer_id
GROUP BY state
) Q
GROUP BY transact_amt
ORDER BY HighestSum
I'm lost. Thank you.
Expected results are the names of customers with the top 5 highest transactions in each state.
ERROR: table name "customers" specified more than once
SQL state: 42712

Answer 1

First, you need for your JOIN to be correct. Second, you want to use window functions:
SELECT ct.*
FROM (SELECT c.customer_id, c.name, c.state, SUM(t.transact_amt) AS total,
ROW_NUMBER() OVER (PARTITION BY c.state ORDER BY SUM(t.transact_amt) DESC) as seqnum
FROM customers c JOIN
transaactions t
ON t.customer_id = c.customer_id
GROUP BY c.customer_id, c.name, c.state
) ct
WHERE seqnum <= 5;
You seem to have several issues with SQL. I would start with understanding aggregation functions. You have a SUM() with the alias HighestSum. It is simply the total per customer.

Answer 2

You can get them using aggregation and then by using the RANK() window function. For example:
select
state,
rk,
customer_name
from (
select
*,
rank() over(partition by state order by total desc) as rk
from (
select
c.customer_id,
c.customer_name,
c.state,
sum(t.transact_amt) as total
from customers c
join transactions t on t.customer_id = c.customer_id
group by c.customer_id
) x
) y
where rk <= 5
order by state, rk

Answer 3

There are two valid answers already. Here's a third:
SELECT *
FROM (
SELECT c.state, c.customer_name, t.*
, row_number() OVER (PARTITION BY c.state ORDER BY t.transact_sum DESC NULLS LAST, customer_id) AS rn
FROM (
SELECT customer_id, sum(transact_amt) AS transact_sum
FROM transactions
GROUP BY customer_id
) t
JOIN customers c USING (customer_id)
) sub
WHERE rn < 6
ORDER BY state, rn;
Major points
When aggregating all or most rows of a big table, it's typically substantially faster to aggregate before the join. Assuming referential integrity (FK constraints), we won't be aggregating rows that would be filtered otherwise. This might change from nice-to-have to a pure necessity when joining to more aggregated tables. Related:
Why does the following join increase the query time significantly?
Two SQL LEFT JOINS produce incorrect result
Add additional ORDER BY item(s) in the window function to define which rows to pick from ties. In my example, it's simply customer_id. If you have no tiebreaker, results are arbitrary in case of a tie, which may be OK. But every other execution might return different results, which typically is a problem. Or you include all ties in the result. Then we are back to rank() instead of row_number(). See:
PostgreSQL equivalent for TOP n WITH TIES: LIMIT "with ties"?
While transact_amt can be NULL (has not been ruled out) any sum may end up to be NULL as well. With an an unsuspecting ORDER BY t.transact_sum DESC those customers come out on top as NULL comes first in descending order. Use DESC NULLS LAST to avoid this pitfall. (Or define the column transact_amt as NOT NULL.)
PostgreSQL sort by datetime asc, null first?