Get the distinct sum of a joined table column

To get the result without subquery, you have to resort to advanced window function trickery:

SELECT sum(count(*))       OVER () AS tickets_count
     , sum(min(a.revenue)) OVER () AS atendees_revenue
FROM   tickets   t
JOIN   attendees a ON a.id = t.attendee_id
GROUP  BY t.attendee_id
LIMIT  1;

sqlfiddle

How does it work?

The key to understanding this is the sequence of events in the query:

aggregate functions -> window functions -> DISTINCT -> LIMIT

More details:

• Best way to get result count before LIMIT was applied

Step by step:

1. I GROUP BY t.attendee_id - which you would normally do in a subquery.

2. Then I sum over the counts to get the total count of tickets. Not very efficient, but forced by your requirement. The aggregate function count(*) is wrapped in the window function sum( ... ) OVER () to arrive at the not-so-common expression: sum(count(*)) OVER ().

   And sum the minimum revenue per attendee to get the sum without duplicates.

   You could also use max() or avg() instead of min() to the same effect as revenue is guaranteed to be the same for every row per attendee.

   This could be simpler if DISTINCT was allowed in window functions, but PostgreSQL has not (yet) implemented this feature. Per documentation:

   Aggregate window functions, unlike normal aggregate functions, do not allow DISTINCT or ORDER BY to be used within the function argument list.

3. Final step is to get a single row. This could be done with DISTINCT (SQL standard) since all rows are the same. LIMIT 1 will be faster, though. Or the SQL-standard form FETCH FIRST 1 ROWS ONLY.

What about a simple division:

 Select count(tickets.*) as tickets_count
 , sum(attendees.revenue) / count(attendees.id) as atendees_revenue
 from tickets LEFT OUTER JOIN attendees ON attendees.id = 
 tickets.attendee_id;

This should handle duplicates, triplicates, etcetera.