• Status: Solved
• Priority: Medium
• Security: Public
• Views: 375

Progressive total

Dear Experts,
I Have a table with the following fields and data
ID    NAME      date          Salary
1     Bob     2004-01-01   10
2     John    2004-01-01   10
3     Bob     2004-02-01   30
4     Helen  2004-02-01    20
5     Helen  2004-03-01    50

I want to see the following by single-query

Bob     2004-01-01   10
John    2004-01-01   10
Helen  2004-01-01    0
Bob     2004-02-01   40
John    2004-02-01   10
Helen  2004-02-01    20
Bob     2004-03-01   40
John    2004-03-01   10
Helen  2004-03-01    70
0
DMStar
1 Solution

Commented:
I think

Select m.NAME, m.date, sum(m1.Salary)
from MyTable m
LEFT OUTER JOIN MyTable m1 ON m1.date <= m.date
GROUP BY m.name, m.date
0

Commented:
can you explain the expeted results a bit pls.   I dont understand where the figures come from
0

Commented:
sorry, forgot a join on id.  Are the id's in your example correct, or maybe the join is supposed to be on name, not id.

Select m.NAME, m.date, sum(m1.Salary)
from MyTable m
LEFT OUTER JOIN MyTable m1 ON m1.date <= m.date AND m.id = m1.id
GROUP BY m.name, m.date
0

Commented:
For each date that appears in the Salaries table?

BTW, avoid using reserved words like "date" are column names.
Select Name, SDate, Sum(case S.Date <= D.Date then Salary else 0 end)
FROM Salaries S CROSS JOIN Select distinct [date] as SDate from Salaries) D
0

Commented:
Select Name, SDate, Sum(case S.Date <= D.Date then Salary else 0 end)
FROM Salaries S CROSS JOIN Select distinct [date] as SDate from Salaries) D
GROUP BY Name, SDate
0

Commented:
Adwiseman's solution will miss out Helen's first record at least. Needs to be a cross product one the dates to ensure each date is represented for each Name.
0

Commented:
Here with exactly your output

Select m.NAME, md.date, sum(ISNULL(m1.Salary, 0))
from (Select distinct name from MyTable) m
CROSS JOIN (Select distinct date from MyTable) md
LEFT OUTER JOIN MyTable m1 ON m1.date <= md.date AND m.name = m1.name
GROUP BY m.name, md.date
order by md.date, m.name

I'm selecting a list of names , and cross joining that with the distinct list of dates.  Goving me a list of all names, with all dates.  Then joining that in with the salart and grouping them.  The order is not the same, but the result set is.
0

Commented:
--my working code

create table MyTable([id] int, [name] varchar(50), [date] smalldatetime, salary int)
GO
insert into MyTable
Select 1, 'Bob', '2004-01-01', 10
insert into MyTable
Select 2, 'John', '2004-01-01', 10
insert into MyTable
Select 1, 'Bob', '2004-02-01', 30
insert into MyTable
Select 3, 'Helen', '2004-02-01', 20
insert into MyTable
Select 3, 'Helen', '2004-03-01', 50
GO
select * from MyTable
GO
Select m.NAME, md.date, sum(ISNULL(m1.Salary, 0))
from (Select distinct name from MyTable) m
CROSS JOIN (Select distinct date from MyTable) md
LEFT OUTER JOIN MyTable m1 ON m1.date <= md.date AND m.name = m1.name
GROUP BY m.name, md.date
order by md.date, m.name
0

Commented:
I would sugest using a different source for the list of all distinct dates.  A function that returns a table given a start and end date, or create a table of all dates say between 01-01-1900 and 01-01-2200
0

Commented:
monosodiumg's code, with typo correction works also.

Here, changed to use my test code.

Select Name, SDate, Sum(case WHEN S.Date <= D.SDate then Salary else 0 end)
FROM MyTable S
CROSS JOIN (Select distinct [date] as SDate from MyTable) D
GROUP BY Name, SDate
0

Author Commented:
Your code works fine
2monosodiumg
doesn't work syntax error

Thanks everybody
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Featured Post

Tackle projects and never again get stuck behind a technical roadblock.