Michal... I must apologise, your suggestion worked a treat!!!
I never realised it was possible to do a join on a table to itself before!
2009/4/15 Michal Politowski :
> On Wed, 15 Apr 2009 15:32:42 +0100, Shug Boabby wrote:
>> I simplified my problem a little too much and now I'm stuck trying to
Oh... and also, A, B, C are in the same table.
2009/4/17 Shug Boabby :
> Life sure would be easier if that were the case Michal, but no... that
> is not the case here. The sum is not a simple sum, it is a sum of all
> elements having a lower or equal A and the same C. This is a
> "cumulative sum"
Life sure would be easier if that were the case Michal, but no... that
is not the case here. The sum is not a simple sum, it is a sum of all
elements having a lower or equal A and the same C. This is a
"cumulative sum" as pointed out by others.
2009/4/15 Michal Politowski :
> On Wed, 15 Apr 2009 1
On Wed, 15 Apr 2009 15:32:42 +0100, Shug Boabby wrote:
> I simplified my problem a little too much and now I'm stuck trying to
> use cumulative_sum(). My schema is not only A, B but also has a C
>
> A B C
> 1 0 1
> 2 1 1
> 3 0 1
> 4 2 1
> 5 1 1
> 1 0 2
> 2 1 2
> 3 0 2
> 4 2 2
> 5 1 2
>
> and I wa
I simplified my problem a little too much and now I'm stuck trying to
use cumulative_sum(). My schema is not only A, B but also has a C
A B C
1 0 1
2 1 1
3 0 1
4 2 1
5 1 1
1 0 2
2 1 2
3 0 2
4 2 2
5 1 2
and I want to be able to do the cumulative sum only when C is the same. E.g.
A funkySumB C
1 0
In response to Shug Boabby :
> Hello all,
>
> I have a table with 2 bigint columns, let's call them A and B. I need
> a query that will allow me to return A alongside the sum of Bs from
> rows where A is less than or equal to this row's A. It is best
> described with some example data, consider th
On Wed, Apr 15, 2009 at 11:09:49AM +0100, Shug Boabby wrote:
> Anyone have any ideas how to do this? I'm able to do it
> programmatically, but it's slow. Optimally I'd like to be able to do
> this in the DB. As you can see, it's a little trickier that the usual
> aggregate function with a GROUP BY
