SELECT users with at least 2 identical products - sql

Question

Users table:
user_id name
1 john
2 mark
3 scott
4 piter
user_products table:
user_id product_id
1 2
1 4
1 5
2 4
2 5
2 7
3 1
3 5
3 4
3 2
4 1
As we see, users 1,2,3 all have products 4 and 5. So how to select users with at least 2 identical products ?

Answer 1

One option is to use a self join:
SELECT
u.user_id,
u.name
FROM user_products up1
INNER JOIN user_products up2
ON up1.product_id = up2.product_id AND
up1.user_id <> up2.user_id
INNER JOIN Users u
ON up1.user_id = u.user_id
GROUP BY
u.user_id,
u.name
HAVING
COUNT(DISTINCT up1.product_id) > 1;
Demo
The idea here is to try to match each of the records for a given user to records from a different user, but having the same product. The aggregation step then checks if a given user still has at least two products after the inner join, which implies that in fact he does have at least two products in common with some other user.
The matching users here are: john, mark, and scott

Answer 2

A self join is the right approach, but I think the right logic is:
SELECT up1.user_id, up2.user_id
FROM user_products up1 JOIN
user_products up2
ON up1.product_id = up2.product_id AND
up1.user_id < up2.user_id
GROUP BY up1.user_id, up2.user_id
HAVING COUNT(DISTINCT up1.product_id) >= 2;
If you want the list of products, you can include array_agg(distinct up1.product_id).

Answer 3

You can try this as well:
select distinct b.user_id from #UserProduct b
join (
select count(1) cnt,product_id from #UserProduct
group by product_id
having count(1) = 3) c
on b.product_id=c.product_id

Answer 4

This time, I hope can help you
Select * into #User From (
Select '1' [user_id], 'john' [user_name] Union All
Select '2' [user_id], 'mark' [user_name] Union All
Select '3' [user_id], 'scott' [user_name] Union All
Select '4' [user_id], 'piter' [user_name]
) A
Select * into #UserProduct From (
Select '1' [user_id], '2' [product_id] union All
Select '1' [user_id], '4' [product_id] union All
Select '1' [user_id], '5' [product_id] union All
Select '2' [user_id], '4' [product_id] union All
Select '2' [user_id], '5' [product_id] union All
Select '2' [user_id], '7' [product_id] union All
Select '3' [user_id], '1' [product_id] union All
Select '3' [user_id], '5' [product_id] union All
Select '3' [user_id], '4' [product_id] union All
Select '3' [user_id], '2' [product_id] union All
Select '4' [user_id], '1' [product_id]
) A
Select U1.[user_id] From (
Select
A.[Product_id] Product_id1,
B.[Product_id] Product_id2
From (
Select [Product_id] From #UserProduct
Group By [Product_id]
) A
Left Join (
Select [Product_id] From #UserProduct
Group By [Product_id]
) B On 1 = 1
Where A.[Product_id] < B.[Product_id]
) Product2
Left Join (
Select [user_id] From #UserProduct
Group By [user_id]
) [User] On 1 = 1
Left Join #UserProduct U1 On U1.[user_id] = [User].[user_id] and U1.Product_id = Product_id1
Left Join #UserProduct U2 On U2.[user_id] = [User].[user_id] and U2.Product_id = Product_id2
Where (U1.[user_id] Is Not Null And U2.[user_id] Is Not Null)
Group By U1.[user_id]
Result :

Answer 5

Hope can help you
Select * into #User From (
Select '1' [user_id], 'john' [user_name] Union All
Select '2' [user_id], 'mark' [user_name] Union All
Select '3' [user_id], 'scott' [user_name] Union All
Select '4' [user_id], 'piter' [user_name]
) A
Select * into #UserProduct From (
Select '1' [user_id], '2' [product_id] union All
Select '1' [user_id], '4' [product_id] union All
Select '1' [user_id], '5' [product_id] union All
Select '2' [user_id], '4' [product_id] union All
Select '2' [user_id], '5' [product_id] union All
Select '2' [user_id], '7' [product_id] union All
Select '3' [user_id], '1' [product_id] union All
Select '3' [user_id], '5' [product_id] union All
Select '3' [user_id], '4' [product_id] union All
Select '3' [user_id], '2' [product_id] union All
Select '4' [user_id], '1' [product_id]
) A
Select U.[user_id], U.[user_name], P.[qty_product]
From #User U
Left Join (
Select
[user_id], Count(*) [qty_product]
From #UserProduct
Group By [user_id]
Having Count(*) > 1
) P On P.[user_id] = U.[user_id]
Result :