Find sum of hours for each date worked - sql

Question

I have a table of timesheet entries set up like this:
id
job_id
employee_id
hours_worked
date_worked
1
1
111
8
2022-10-01
2
1
222
8
2022-10-01
3
1
222
8
2022-10-02
4
2
222
8
2022-10-03
5
2
111
8
2022-10-04
6
2
222
5
2022-10-05
7
3
111
8
2022-10-04
8
4
333
8
2022-10-07
9
4
111
3
2022-10-09
I'm trying to find the sum of hours for the first, second, third etc dates that work was done on each job
Ideally I'd like something like this:
job_id
Day1_hours
Day2_hours
Day3_hours
1
16
8
0
2
8
8
5
3
8
0
0
4
8
3
0
The trouble I'm running into is that there can be multiple employees working on each day of the job, so using a query to select the min(date_worked) greater than a subquery for min(date_worked) is sometimes giving me the same dates. There are sometimes days in between work done on a job, so I can't just add a day to the minimum value and check hours for that date.
How can I find the sum of hours_worked for the first date_worked, then the second, third etc?

Answer 1

PIVOT's are great but conditional aggregations offer a bit more flexibility
Example
Select job_id
,[Day1_Hours] = sum( case when DN=1 then hours_worked else 0 end)
,[Day2_Hours] = sum( case when DN=2 then hours_worked else 0 end)
,[Day3_Hours] = sum( case when DN=3 then hours_worked else 0 end)
From ( Select *
,DN = dense_rank() over (partition by job_id order by date_worked)
From YourTable
) A
Group By Job_ID

Answer 2

select job_id
,[1] as day1_hours
,[2] as day2_hours
,[3] as day3_hours
from (
select job_id
,hours_worked
,dense_rank() over(partition by job_id order by date_worked) as days
from t
) t
pivot (sum(hours_worked) for days in([1],[2],[3])) p
job_id
day1_hours
day2_hours
day3_hours
1
16
8
null
2
8
8
5
3
8
null
null
4
8
3
null
Fiddle