> -----Original Message-----
> From: pgsql-sql-owner@postgresql.org [mailto:pgsql-sql-
> owner@postgresql.org] On Behalf Of air
> Sent: Saturday, October 06, 2012 8:48 AM
> To: pgsql-sql@postgresql.org
> Subject: [SQL] How to make this CTE also print rows with 0 as count?
>
> I have a CTE based query, to which I pass about 2600 4-tuple
> latitude/longitude values using joins - these latitude longitude 4-tuples
have
> been ID tagged and held in a second table called coordinates. These top
left
> and bottom right latitude / longitude values are passed into the CTE in
order
> to display the amount of requests (hourly) made within those coordinates
> for given two timestamps).- I am able to get the total requests per day
within
> the timestamps given, that is, the total count of user requests on every
> specified day. (E.g. user opts to see every Wednesday or Wednesday AND
> Thursday etc. - between hours 11:55 and 22:04 between dates January 1 and
> 31, 2012 for every latitude/longitude 4-tuples I pass.) But I cannot view
the
> rows with count 0. My query is as below:
>
>
>
> WITH v AS (
> SELECT '2012-01-1 11:55:11'::timestamp AS _from
> ,'2012-01-31 22:02:21'::timestamp AS _to
> )
> , q AS (
> SELECT c.coordinates_id
> , date_trunc('hour', t.calltime) AS stamp
> , count(*) AS zcount
> FROM v
> JOIN mytable t ON t.calltime BETWEEN v._from AND v._to
> AND (t.calltime::time >= v._from::time AND
> t.calltime::time <= v._to::time) AND (extract(DOW
from
> t.calltime) = 3)
> JOIN coordinates c ON (t.lat, t.lon)
> BETWEEN (c.bottomrightlat, c.topleftlon)
> AND (c.topleftlat, c.bottomrightlon)
> GROUP BY c.coordinates_id, date_trunc('hour', t.calltime)
> )
> , cal AS (
> SELECT generate_series('2011-2-2 00:00:00'::timestamp
> , '2012-4-1 05:00:00'::timestamp
> , '1 hour'::interval) AS stamp
> FROM v
> )
> SELECT q.coordinates_id, cal.stamp::date, sum(q.zcount) AS zcount
> FROM v, cal
> LEFT JOIN q USING (stamp)
> WHERE extract(hour from cal.stamp) >= extract(hour from v._from)
> AND extract(hour from cal.stamp) <= extract(hour from v._to)
> AND extract(DOW from cal.stamp) = 3
> AND cal.stamp >= v._from
> AND cal.stamp <= v._to
> GROUP BY q.coordinates_id, cal.stamp::date ORDER BY q.coordinates_id,
> stamp;
>
>
>
>
> The output I get when I execute this query is basically like this
(normally I
> have about 10354 rows returned excluding the rows with 0 zcount, just
> providing two coordinates for sake of similarity):
>
> coordinates_id | stamp | zcount
> 1 ;"2012-01-04"; 2
> 1 ;"2012-01-11"; 3
> 1 ;"2012-01-18"; 2
> 2 ;"2012-01-04"; 2
> 2 ;"2012-01-11"; 3
> 2 ;"2012-01-18"; 2
>
>
>
>
> However, it should be like this where all rows with zcount 0 should also
be
> printed out along with rows that have nonzero zcounts -E.g. January 25
with
> zcount 0 for the two coordinates with ID 1 and 2 should also be printed in
this
> small portion of example-:
>
> coordinates_id | stamp | zcount
> 1 ;"2012-01-04"; 2
> 1 ;"2012-01-11"; 3
> 1 ;"2012-01-18"; 2
> 1 ;"2012-01-25"; 0
> 2 ;"2012-01-04"; 2
> 2 ;"2012-01-11"; 3
> 2 ;"2012-01-18"; 2
> 2 ;"2012-01-25"; 0
>
>
>
>
> How can I achieve this? Thanks in advance.
>
>
Food for thought, generally when you want "everything including the zeros"
you want to build the "master" set without any values, build out the "values
only" dataset, then LEFT JOIN them and use COALESCE to generate values for
the missing data.
So:
SELECT id, stamp, COALESCE(datavalues.zcount, 0) AS zcount
FROM (cal CROSS JOIN id_master) master
LEFT JOIN datavalues USING (id, stamp)
Also, the mixing of multiple FROM relations and JOINs is confusing. In
particular is the fact the JOIN takes precedence over the "," in FROM
"A JOIN clause combines two FROM items. Use parentheses if necessary to
determine the order of nesting. In the absence of parentheses, JOINs nest
left-to-right. In any case JOIN binds more tightly than the commas
separating FROM items."
http://www.postgresql.org/docs/9.2/interactive/sql-select.html
Your query is equivalent to:
SELECT ...
FROM v CROSS JOIN (cal LEFT JOIN q USING stamp)
WHERE ...
Anyway, the "create master, left join data, coalesce" methodology is one
that I find to be easy to understand and implement.
HTH,
David J.
