Question

Continuing on my last question…

Let me try to explain my schema. I have three tables we’ll call users (with columns id and name), parties (with columns id, partydate, and user_id) and questions (with columns id, createdate, and user_id). My requirement is to show for every user the number of parties within the last year and questions created within the last year.

My query looks like:

SELECT users.id, users.name,  
  SUM(CASE WHEN (parties.party> NOW() - interval '1 year') THEN 1 ELSE 0 END) 
    AS numparties, 
  SUM(CASE WHEN (questions.createdate> NOW() - interval '1 year') THEN 1 ELSE 0 END)
    AS numquestions
FROM users
  LEFT JOIN parties ON users.id=parties.user_id
  LEFT JOIN questions ON users.id=questions.user_id
GROUP BY users.id, users.name;

This works almost 100%. I am getting a result with all users that exist. The problem is, for some users (a very small few) I’m counting either a party or a question twice. For example, if I change the above query to just show parties.id and questions.id instead of summing them as well as remove the GROUP BY, I might get something like:

user.id | user.name | parties.id | questions.id  
-----------------------------------------------
0          John          15             2
0          John          15             7

You can see it shows the parties.id twice.

When I was using COUNT() I could rely on DISTINCT but with SUM I’m not sure how I can. I want something like:

SUM(CASE WHEN (parties.party> NOW() - interval '1 year' AND parties.id IS DISTINCT) THEN 1 ELSE 0 END) 
AS numparties, 

But of course this isn’t valid. Can this small problem be corrected easily?

Answer 1

I won’t write the code for you (since it’s homework), but you’ll want to put the two calculations into subqueries.

Here’s a template:

  SELECT users.id, users.name, 
         subquery1.result_of_calculation1, subquery2.result_of_calculation2
    FROM users
         LEFT JOIN (
            --calculation 1
         ) subquery1
         ON users.id = subquery1.user_id
         LEFT JOIN (
            --calculation 2
         ) subquery2
         ON users.id = subquery2.user_id
GROUP BY users.id, users.name;