I have a problem here, and I\'m hoping there is an easy solution. I\'ll try to make this as simple as possible:
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
The key to understanding this is the sequence of events in the query:
aggregate functions -> window functions -> DISTINCT -> LIMIT
More details:
Step by step:
I GROUP BY t.attendee_id
- which you would normally do in a subquery.
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
orORDER BY
to be used within the function argument list.
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.
Previous answer is nearly correct. You just need to make distinct work well in case identical revenues. You can fix this really simple if your id has numeric type:
SELECT
COUNT(*) AS ticketsCount,
SUM(DISTINCT attendees.id + attendees.revenue) - SUM(DISTINCT attendees.id) AS revenueSum
FROM
tickets
LEFT JOIN attendees ON
attendees.id = tickets.attendee_id
You were actually pretty close, there's many way to do this and if I understand your question correctly this should do it :
SELECT
COUNT(*) AS ticketsCount,
SUM(DISTINCT attendees.revenue) AS revenueSum
FROM
tickets
LEFT JOIN attendees ON
attendees.id = tickets.attendee_id