Solutions that do the job "in a minimum time" are always challenging. Success depends on the existence of proper indexes. Assuming these are in place, we can still sometimes see markedly different results for solutions written with different query constructions.
The first step towards a solution, the most important step, is to make sure we understand the exact requirements. In this case, we don't care about purchase data, just that it exists. We aren't actually retrieving anything from the PURCHASES table! If you had said "give the date of the latest Product X purchase, and the total number of Product Y purchases" then we'd need to write a totally different query.
Here's one solution:
select customer_id from CUSTOMERS as C where exists ( select * from PURCHASES where customer_id = C.customer_id and product_id = 'X' ) and 2 <= ( select count(*) from PURCHASES where customer_id = C.customer_id and product_id = 'Y' and purchase_date between date1 and date2 )
Each of the two subqueries above is a correlated subquery. This means that it considers only those purchases which match the customer_id of the correlated row in the main query.
One advantage of using correlated subqueries is that it's fairly easy to understand what they're doing. They "read" well. In this case, though, there are two of them, which leaves open the possibility that the database optimizer will generate two separate joins in order to execute them. (Correlated subqueries are usually executed as joins.)
Here's a different solution:
select C.customer_id from CUSTOMERS as C inner join PURCHASES as P on C.customer_id = P.customer_id group by C.customer_id having 0 < sum( case when P.product_id = 'X' then 1 else 0 end ) and 2 <= sum( case when P.product_id = 'Y' and P.purchase_date between date1 and date2 then 1 else 0 end )
Here you can see we've taken matters into our own hands and performed one join explicitly. Note that we're still just selecting from the CUSTOMERS table. The WHERE EXISTS construction is replaced by taking a count and making sure it's not zero. The counts are achieved by obtaining the SUM of a column of 1's and 0's.
Which of the solutions is faster? Try them both, and see.
This was first published in March 2005