COUNT all values in a column with JOIN - mysql

Question

I am joining three tables and need to return two separate counts, one showing the total number of unique users who have purchased an item, and the other showing the total number of unique users who haven't purchased an item. These are cropped for brevity, but here are the relevant tables:
user table
+----------+------+------+-----+
| username | colb | colc | etc |
+----------+------+------+-----+
| user1 | * | * | * |
| user2 | * | * | * |
| user3 | * | * | * |
+----------+------+------+-----+
purchase table
+------------+---------+----------+------+
| purchaseID | storeID | username | cost |
+------------+---------+----------+------+
| 1 | 1 | user1 | * |
| 2 | 1 | user2 | * |
| 3 | 5 | user2 | * |
| 4 | 3 | user1 | * |
+------------+---------+----------+------+
store table
+---------+-----------+-----+
| storeID | storeName | etc |
+---------+-----------+-----+
| 1 | store1 | * |
| 2 | store2 | * |
| 3 | store3 | * |
+---------+-----------+-----+
I am currently using this query to get the unique users who have purchased an item from a store:
SELECT
store.storeID storeID,
store.storeName storeName,
COUNT(DISTINCT CASE WHEN purchase.username IS NOT NULL
THEN purchase.purchaseID END) AS purchases
[Query to retrieve total unique users who have not purchased an item]
FROM store
LEFT JOIN purchase
ON store.storeID = purchase.storeID
LEFT JOIN user
ON purchase.username = user.username
GROUP BY 1, 2
I have tried a few different ways, none of which have worked. The issue I've identified is when the LEFT JOIN happens it only returns the matching results for usernames, thus the COUNT won't include the other users in the user table. I have not had any luck finding a way to fix this, so I'm hoping someone on here can lend me a hand. The results I'm hoping to see should be something like this:
+---------+-----------+-----------+--------------+
| storeID | storeName | purchases | nonPurchases |
+---------+-----------+-----------+--------------+
| 1 | store1 | 2 | 1 |
| 2 | store2 | 0 | 3 |
| 3 | store3 | 1 | 2 |
+---------+-----------+-----------+--------------+

Answer 1

that is actually quite simple.
First you count all user and subtract te count of distinct purchasers
SELECT
store.storeID storeID,
store.storeName storeName,
COUNT(DISTINCT CASE WHEN purchase.username IS NOT NULL
THEN purchase.purchaseID END) AS purchases,
(SELECT COUNT(*) FROM User) - COUNT(DISTINCT CASE WHEN purchase.username IS NOT NULL
THEN purchase.purchaseID END) AS NON_purchases
FROM store
LEFT JOIN purchase
ON store.storeID = purchase.storeID
LEFT JOIN user
ON purchase.username = user.username
GROUP BY 1, 2

Answer 2

Here is a clean solution.
Please note that the aggregation is done before the join.
with
purchases as
(
select storeID
,count(distinct username) as purchase
from purchase
group by storeID
),
users as
(
select count(*) as total_users
from user
)
select storeID
,storeName
,coalesce(purchase, 0) as purchase
,total_users - coalesce(purchase, 0) as nonPurchases
from store
left join purchases using (storeID)
cross join users
storeID
storeName
purchase
nonPurchases
1
store1
2
1
2
store2
0
3
3
store3
1
2
Fiddle

Answer 3

I'll go with a slightly different approach.
Generate a combination of store and user using CROSS JOIN, make it as a subquery then use that to LEFT JOIN with purchase table. In SELECT, change COUNT(DISTINCT ..) to SUM(..). Something like this:
SELECT us.storeID,
us.storeName,
SUM(CASE WHEN p.username IS NOT NULL
THEN 1 ELSE 0 END) AS purchases,
SUM(CASE WHEN p.username IS NULL
THEN 1 ELSE 0 END) AS nonPurchases
FROM (SELECT storeID, storeName, username FROM user u CROSS JOIN store s) us
LEFT JOIN (SELECT DISTINCT storeid, username FROM purchase) p
ON us.storeID = p.storeID
AND us.username=p.username
GROUP BY 1, 2;
Thanks to David pointing out in the comment that my previous suggestion is not exactly counting unique users. So I made a quick modification to make sure that it does what OP wanted in the first place. Therefore I did a SELECT DISTINCT ... on purchase table then make it as a subquery for the LEFT JOIN. The other parts of the original suggestion remains.
Updated fiddle