Do different sums in the same query - sql

Question

I currently have an issue on a query:
I have 2 tables.
Table 1:
Table 2:
I'm trying to join both tables on DateHour (that works) and for each campaign, for each PRF_ID, for each LENGTH and for each Type, to calculate count the occurences of the HPP column of table 2, year per year.
So for instance, for a given PRF_ID, length,type and campaign, I will have a range of dates in Table 1 between 01/04/2019 and 01/04/2020.
In this case, in my query, I need a new column giving me for all the dates between 01/04/2019 and 31/12/2019 the sum of HPP ocurrences in this period.
For 2020, the sum would be between 01/01/2020 and 01/04/2020.
I tried doing something like this:
SELECT Table1.DateHour,
SUM(Table2.HPP) OVER (PARTITION BY YEAR(Table2.DateHour)
FROM Table1
LEFT JOIN Table2 on Table2.DateHour=Table1.DateHour
But that gives me really odd results, the OVER PARTITION BY does not seem to work.

Answer 1

Your question is confusing because it mixes terminology.
Count versus sum
... to calculate count the occurences ... the sum would be ...
Counting occurrences is not the same as adding them up. Every records that can be joined counts as an occurence. Calculating the sum means adding up the values of a column. I added both calculations to the solution below (see IsHour_Sum versus Table2_Count).
Grouping on combinations
Do different sums in the same query [question title]
... and for each campaign, for each PRF_ID, for each LENGTH and for each Type ...
Do you want to aggregate over each combination of those columns or do you want to aggregate over each column individually? I have assumed you are after the combinations in my solution. Example to clarify:
If
column A has 3 possible values (A1, A2, A3) and
column B has 2 possible values (B1, B2)
Then
there are 5 counts (3 + 2) when aggregating (A) and (B) indivually
there are 6 counts (3 * 2) when aggregating each combination of (A,B)
Again:
A1 -> count 1 vs A1,B1 -> count 1
A2 -> count 2 A1,B2 -> count 2
A3 -> count 3 A2,B1 -> count 3
B1 -> count 4 A2,B2 -> count 4
B2 -> count 5 A3,B1 -> count 5
A3,B2 -> count 6
Sample data
I left out the column Value from table1 because it is not part of your question. I also changed the dates for table2 to 2018-05-23 to match with the majority of the table1 records. Otherwise all counts and sums would be 0.
declare #table1 table
(
DateHour datetime,
PRF_ID int,
Campaign nvarchar(5),
Length int,
ContractType nvarchar(5)
);
insert into #table1 (DateHour, PRF_ID, Campaign, Length, ContractType) values
('2018-05-23 00:00', 1, 'Q218', 1, 'G'),
('2018-05-23 01:00', 1, 'Q218', 1, 'G'),
('2018-05-23 02:00', 1, 'Q218', 1, 'G'),
('2020-05-23 03:00', 1, 'Q120', 1, 'G'),
('2018-05-23 04:00', 1, 'Q218', 1, 'G'),
('2019-07-23 01:00', 1, 'Q219', 1, 'G');
declare #table2 table
(
DateHour datetime,
HPP int
);
insert into #table2 (DateHour, HPP) values
('2018-05-23 00:00', 0),
('2018-05-23 01:00', 0),
('2018-05-23 02:00', 1),
('2018-05-23 03:00', 0),
('2018-05-23 04:00', 1),
('2018-05-23 05:00', 0),
('2018-05-23 06:00', 0),
('2018-05-23 07:00', 0);
Solution
The easiest way to aggregation on the year of the dates, is to split of that part as a new column instead of using a over(partition by ...) construction. If you do not need the new column Year in your output, then you can simply remove it from the field list (after select), but it must remain in the grouping clause (after group by).
select year(t1.DateHour) as 'Year',
t1.PRF_ID,
t1.Campaign,
t1.Length,
t1.ContractType,
isnull(sum(t2.HPP), 0) as 'IsHour_Sum',
count(t2.DateHour) as 'Table2_Count'
from #table1 t1
left join #table2 t2
on t2.DateHour = t1.DateHour
/* -- specify date filter as required
where t1.DateHour >= '2019-04-01 00:00'
and t1.DateHour < '2020-04-01 00:00'
*/
group by year(t1.DateHour),
t1.PRF_ID,
t1.Campaign,
t1.Length,
t1.ContractType;
Result
Year PRF_ID Campaign Length ContractType IsHour_Sum Table2_Count
----------- ----------- -------- ----------- ------------ ----------- ------------
2018 1 Q218 1 G 2 4
2019 1 Q219 1 G 0 0
2020 1 Q120 1 G 0 0