help on sql

Hi sql gurus,

I need some help and also curious on how to write this in single sql
statement, requirements goes like this

Table structure
================
create table find_fixed_open (
find number,
fixed number,
open number,
bug_when date
);


Data
=============
insert into find_fixed_open values(5,3,2,'10/01/2009');
insert into find_fixed_open values(52,38,16,'10/02/2009');
insert into find_fixed_open values(68,45,39,'10/03/2009');
insert into find_fixed_open values(112,59,92,'10/04/2009');
insert into find_fixed_open values(45,12,125,'10/05/2009');


Formula
=========
Open = Find - fixed + previous rows open(basically open is cumulative)
i.e.
5 - 3 + 0 = 2
52 - 38 + 2 = 16
68 - 45 + 16 = 39
112 - 59 + 39 = 92
45 - 12 + 92 = 125

Please help me and enlighten me.

Thank you
Raju

On Oct 9, 10:39 pm, RA wrote:
> Hi sql gurus,
>
> I need some help and also curious on how to write this in single sql
> statement, requirements goes like this
>
> Table structure
> ================
> create table find_fixed_open (
> find number,
> fixed number,
> open number,
> bug_when date
> );
>
> Data
> =============
> insert into find_fixed_open values(5,3,2,'10/01/2009');
> insert into find_fixed_open values(52,38,16,'10/02/2009');
> insert into find_fixed_open values(68,45,39,'10/03/2009');
> insert into find_fixed_open values(112,59,92,'10/04/2009');
> insert into find_fixed_open values(45,12,125,'10/05/2009');
>
> Formula
> =========
> Open = Find - fixed + previous rows open(basically open is cumulative)
> i.e.
> 5 - 3 + 0 = 2
> 52 - 38 + 2  = 16
> 68 - 45 + 16 = 39
> 112 - 59 + 39 = 92
> 45 - 12  + 92 = 125

So is this the result you want:
2
16
39
92
125
??????

or a single summand, as in
294
??????????

(Hint to other posters: clear descriptions of your problem go a long
way to getting help)

>
> Please help me and enlighten me.
>
> Thank you
> Raju

Aside from OPEN being a keyword,

Break the problem down into steps.
The individual counts can be obtained using
SELECT (find-fixed+open) opencount from find_fixed_open;

Assuming you wanted the individual counts, you can stop here.

That basically gives you a table of the individual counts. Now given
that table, how would you compute the sum? Here's a hint:
SELECT SUM(opencount) ...

Assuming you want the single summation, can you finish the work by
putting these hints together into one statement?

HTH,
Ed