COUNT with multiple LEFT joins [duplicate] - sql

Question

This question already has answers here:
Two SQL LEFT JOINS produce incorrect result
(3 answers)
Closed 12 months ago.
I am having some troubles with a count function. The problem is given by a left join that I am not sure I am doing correctly.
Variables are:
Customer_name (buyer)
Product_code (what the customer buys)
Store (where the customer buys)
The datasets are:
Customer_df (list of customers and product codes of their purchases)
Store1_df (list of product codes per week, for Store 1)
Store2_df (list of product codes per day, for Store 2)
Final output desired:
I would like to have a table with:
col1: Customer_name;
col2: Count of items purchased in store 1;
col3: Count of items purchased in store 2;
Filters: date range
My query looks like this:
SELECT
DISTINCT
C_customer_name,
C.product_code,
COUNT(S1.product_code) AS s1_sales,
COUNT(S2.product_code) AS s2_sales,
FROM customer_df C
LEFT JOIN store1_df S1 USING(product_code)
LEFT JOIN store2_df S2 USING(product_code)
GROUP BY
customer_name, product_code
HAVING
S1_sales > 0
OR S2_sales > 0
The output I expect is something like this:
Customer_name
Product_code
Store1_weekly_sales
Store2_weekly_sales
Luigi
120012
4
8
James
100022
6
10
But instead, I get:
Customer_name
Product_code
Store1_weekly_sales
Store2_weekly_sales
Luigi
120012
290
60
James
100022
290
60
It works when instead of COUNT(product_code) I do COUNT(DSITINCT product_code) but I would like to avoid that because I would like to be able to aggregate on different timespans (e.g. if I do count distinct and take into account more than 1 week of data I will not get the right numbers)
My hypothesis are:
I am joining the tables in the wrong way
There is a problem when joining two datasets with different time aggregations
What am I doing wrong?

Answer 1

The reason as Philipxy indicated is common. You are getting a Cartesian result from your data thus bloating your numbers. To simplify, lets consider just a single customer purchasing one item from two stores. The first store has 3 purchases, the second store has 5 purchases. Your total count is 3 * 5. This is because for each entry in the first is also joined by the same customer id in the second. So 1st purchase is joined to second store 1-5, then second purchase joined to second store 1-5 and you can see the bloat. So, by having each store pre-query the aggregates per customer will have AT MOST, one record per customer per store (and per product as per your desired outcome).
select
c.customer_name,
AllCustProducts.Product_Code,
coalesce( PQStore1.SalesEntries, 0 ) Store1SalesEntries,
coalesce( PQStore2.SalesEntries, 0 ) Store2SalesEntries
from
customer_df c
-- now, we need all possible UNIQUE instances of
-- a given customer and product to prevent duplicates
-- for subsequent queries of sales per customer and store
JOIN
( select distinct customerid, product_code
from store1_df
union
select distinct customerid, product_code
from store2_df ) AllCustProducts
on c.customerid = AllCustProducts.customerid
-- NOW, we can join to a pre-query of sales at store 1
-- by customer id and product code. You may also want to
-- get sum( SalesDollars ) if available, just add respectively
-- to each sub-query below.
LEFT JOIN
( select
s1.customerid,
s1.product_code,
count(*) as SalesEntries
from
store1_df s1
group by
s1.customerid,
s1.product_code ) PQStore1
on AllCustProducts.customerid = PQStore1.customerid
AND AllCustProducts.product_code = PQStore1.product_code
-- now, same pre-aggregation to store 2
LEFT JOIN
( select
s2.customerid,
s2.product_code,
count(*) as SalesEntries
from
store2_df s2
group by
s2.customerid,
s2.product_code ) PQStore2
on AllCustProducts.customerid = PQStore2.customerid
AND AllCustProducts.product_code = PQStore2.product_code
No need for a group by or having since all entries in their respective pre-aggregates will result in a maximum of 1 record per unique combination. Now, as for your needs to filter by date ranges. I would just add a WHERE clause within each of the AllCustProducts, PQStore1, and PQStore2.