Counting number of positive value in a query - sql

Question

I'm working on the following query and table
SELECT dd.actual_date, dd.week_number_overall, sf.branch_id, AVG(sf.overtarget_qnt) AS targetreach
FROM sales_fact sf, date_dim dd
WHERE dd.date_id = sf.date_id
AND dd.week_number_overall BETWEEN 88-2 AND 88
AND sf.branch_id = 1
GROUP BY dd.actual_date, branch_id, dd.week_number_overall
ORDER BY dd.actual_date ASC;
ACTUAL_DATE WEEK_NUMBER_OVERALL BRANCH_ID TARGETREACH
----------- ------------------- ---------- -----------
13/08/14 86 1 -11
14/08/14 86 1 12
15/08/14 86 1 11.8
16/08/14 86 1 1.4
17/08/14 86 1 -0.2
19/08/14 86 1 7.2
20/08/14 87 1 16.6
21/08/14 87 1 -1.4
22/08/14 87 1 14.4
23/08/14 87 1 2.8
24/08/14 87 1 18
26/08/14 87 1 13.4
27/08/14 88 1 -1.8
28/08/14 88 1 10.6
29/08/14 88 1 7.2
30/08/14 88 1 14
31/08/14 88 1 9.6
02/09/14 88 1 -3.2
the "TargetReach" column shows whether target has been reach or not.
A negative value means target wasn't reached on that day.
How can I get calculate the number of ROW with positive value for this query?
that will show something like:
TOTAL_POSITIVE_TARGET_REACH WEEK_NUMBER_OVERALL
--------------------------- ------------------
13 88
I have tried to use CASE but still not working right.
Thanks a lot.

Answer 1

You want to use conditional aggregation:
with t as (
<your query here>
)
select week_number_overall, sum(case when targetreach > 0 then 1 else 0 end)
from t
group by week_number_overall;
However, I would rewrite your original query to use proper join syntax. Then the query would look like:
SELECT week_number_overall,
SUM(CASE WHEN targetreach > 0 THEN 1 ELSE 0 END)
FROM (SELECT dd.actual_date, dd.week_number_overall, sf.branch_id, AVG(sf.overtarget_qnt) AS targetreach
FROM sales_fact sf JOIN
date_dim dd
ON dd.date_id = sf.date_id
WHERE dd.week_number_overall BETWEEN 88-2 AND 88 AND sf.branch_id = 1
GROUP BY dd.actual_date, branch_id, dd.week_number_overall
) t
GROUP BY week_number_overall
ORDER BY week_number_overall;
THe difference between a CTE (the first solution) and a subquery is (in this case) just a matter of preference.

Answer 2

SELECT WEEK_NUMBER_OVERALL, COUNT(*) TOTAL_POSITIVE_TARGET_REACH
FROM (your original query)
WHERE TARGETREACH >= 0
GROUP BY WEEK_NUMBER_OVERALL

Answer 3

select sum( decode( sign( TARGETREACH ) , -1 , 0 , 0 , 0 , 1 , 1 ) )
from ( "your query here" );

Answer 4

Use HAVING Clause
SELECT dd.actual_date, dd.week_number_overall, sf.branch_id, AVG(sf.overtarget_qnt) AS targetreach
FROM sales_fact sf, date_dim dd
WHERE dd.date_id = sf.date_id
AND dd.week_number_overall BETWEEN 88-2 AND 88
AND sf.branch_id = 1
GROUP BY dd.actual_date, branch_id, dd.week_number_overall
HAVING AVG(sf.overtarget_qnt)>0
ORDER BY dd.actual_date ASC;

Answer 5

Using decode(), sign() get both positive count & negative count.
drop table test;
create table test (
key number(5),
value number(5));
insert into test values ( 1, -9 );
insert into test values ( 2, -8 );
insert into test values ( 3, 10 );
insert into test values ( 4, 12 );
insert into test values ( 5, -9 );
insert into test values ( 6, 8 );
insert into test values ( 7, 51 );
commit;
select sig , count ( sig ) from
(
select key, ( (decode( sign( value ) , -1 , '-ve' , 0 , 'zero' , 1 , '+ve' ) ) ) sig
from test
)
group by sig
SIG COUNT(SIG)
---- ----------------------
+ve 4
-ve 3