SQL: Count distinct column combinations with multiple conditions - sql

Question

I have a joined table that looks like this where mytable contains date, storeid, item and units, whereas stores has state and storeid.
date state storeid item units
==============================================
2020-01-22 new york 712 a 5
2020-01-22 new york 712 b 7
2020-02-18 new york 712 c 0
2020-05-11 new york 518 b 9
2020-01-22 new york 518 b 10
2020-01-21 oregon 613 b 0
2020-02-13 oregon 613 b 9
2020-04-30 oregon 613 b 10
2020-01-22 oregon 515 c 3
And I am trying to create a column that counts the unique number of times that both storeid and item occur in a row where the date is between a given date range and units is greater than 0. Also, only need to select/group by state and the calculated column. I have something that looks like this:
select
s.state,
count(distinct case
when m.date between '2020-01-01' and '2020-03-31'
and m.units > 0
then m.storeid, m.item
end) as q1_total
from mytable as m
left join (select
state,
storeid
from stores) s
on m.storeid=s.storeid
group by s.state
I know my count function isn't written correctly, but I'm not sure how to fix it. Trying to get the end result to look like this.
state q1total
=========================
new york 3
oregon 2

Answer 1

With this query:
select distinct state, storeid, item
from mytable
where date between '2020-01-01' and '2020-03-31' and units > 0
you get all the distinct combinations of state, storeid and item under your conditions and you can aggregate on it:
select state, count(*) q1total
from (
select distinct state, storeid, item
from mytable
where date between '2020-01-01' and '2020-03-31' and units > 0
) t
group by state
See the demo.
Results:
> state | q1total
> :------- | ------:
> new york | 3
> oregon | 2