SELECT sum(played::integer), sum(stats_exists::integer)
FROM matches
WHERE origin=1 AND stats_exists AND NOT training AND match_date > '2009-08-01';
Your approach is elegant, Thom.
But if it is to do it that way then I guess
you should drop the "AND stats_exists" part of the query because
it will filter out every line with stats_exists == f, and those (occasionally) played== t won't get summed up by the SUM() function,
ain't I right ?
Best,
Oliveiros
----- Original Message -----
Sent: Friday, November 20, 2009 2:30 PM
Subject: Re: [NOVICE] Counting booleans for two columns
2009/11/20 Rikard Bosnjakovic
<rikard.bosnjakovic@gmail.com>I have a table with values similiar to this (I have reduced the number
of rows from the actual table):
played | stats_exists
--------+--------------
t | t
t | f
t | t
t | t
f | t
t | t
t | f
What I want to do is to count the number of "t" in each column and
return the values (in two separate columns). The only thing I've
managed to do is doing a UNION, but this gives me the (correct)
results in one column only, I want the results in two (I need to
distinguish between the numbers). I did like this:
SELECT count(played) AS played
FROM matches
WHERE origin=1 AND played AND NOT training AND match_date > '2009-08-01'
UNION
SELECT count(stats_exists) AS stats
FROM matches
WHERE origin=1 AND stats_exists AND NOT training AND match_date > '2009-08-01';
with this result:
played
--------
12 <-- stats
13 <-- played
(2 rows)
How can I rewrite the SELECT so I get two values in separate columns
instead of two rows?
Try:
SELECT sum(played::integer), sum(stats_exists::integer)
FROM matches
WHERE origin=1 AND stats_exists AND NOT training AND match_date > '2009-08-01';
If you're using an old version of PostgreSQL, you might have to use:
SELECT sum(case played when true then 1 else 0 end), sum(case stats_exists when true then 1 else 0 end)
FROM matches
WHERE origin=1 AND stats_exists AND NOT training AND match_date > '2009-08-01';
Regards
Thom