RE: SQL Sum(Distinct())

Actually, the way I interpret the code, there is probably only one header
record per order. If that's true, GROUP BY will allow him to drop the SUM()
from the order header record, and the aggregate issue won't apply. But
we're left to infer, rather than understand, how the data is organized.

Dennis Lovelady
http://www.linkedin.com/in/dennislovelady
--
"When a man says he approves of something in principle, it means he hasn't
the slightest intention of putting it into practice."
-- Prince Otto von Bismark

Using aggregate functions across different join levels of data is
always doing to be a problem...

You could use a Common Table Expression to bring the data to the same
level..

with Hdr as (select CUSNR, ROUTE, STOPX, SHPDTISO
SUM(ORDAM) as SumOrdAmt
from ORHDR
where CUSNR = :CUSNR and
ROUTE = :ROUTE and
STOPX = :STOPX and
SHPDTISO = :SHPDTISO)
, dtl as (select CUSNR, ROUTE, STOPX, SHPDTISO
SUM(ITNSA) as SumItmAmt
from ORDTL
where CUSNR = :CUSNR and
ROUTE = :ROUTE and
STOPX = :STOPX and
SHPDTISO = :SHPDTISO)
select SumOrdAmt, SumItmAmt into :ORDAM, :ITNSA
from hdr join dtl using(CUSNR, ROUTE, STOPX, SHPDTISO)


An alternative format that might be easier to understand

with OrderSummary as (select h.ONRCU, h.ORDAM,
sum(d.ITNSA) as SumItmAmt
from ORHDR H join ORDTL D using
(ONRCU)
Where h.CUSNR = :CUSNR and
h.ROUTE = :ROUTE and
h.STOPX = :STOPX and
h.SHPDTISO = :SHPDTISO
group by h.onrcu)
select sum(ORDAM), sum(SumItmAmt) into :ORDAM, :ITNSA
from OrderSummary

HTH,
Charles


On Mon, Jun 21, 2010 at 10:32 AM, Jeff Crosby
<jlcrosby@xxxxxxxxxxxxxxxx> wrote:

Been a while since I've asked an SQL question, so I'm due.  :)

I will soon, I think, have a need within an RPG program to get a

couple of

pieces of summary information from an order header/detail pair of

files.

One piece from the header file and one piece from the detail file.  I

can

get the 2 pieces separately like this:

        Exec SQL
          Select SUM(ORDAM)
            Into :ORDAM
            From ORHDR
            Where CUSNR = :CUSNR and
                  ROUTE = :ROUTE and
                  STOPX = :STOPX and
                  SHPDTISO = :SHPDTISO;

        Exec SQL
          Select SUM(d.ITNSA)
            Into :ITNSA,
            From ORDTL d Join ORHDR h on d.ONRCU = h.ONRCU
            Where h.CUSNR = :CUSNR and
                  h.ROUTE = :ROUTE and
                  h.STOPX = :STOPX and
                  h.SHPDTISO = :SHPDTISO;


I wondered if I could get the 2 pieces with a single statement, so I

tried

this:

        Exec SQL
          Select SUM(d.ITNSA),
                 SUM(h.ORDAM)
            Into :ITNSA,
                 :ORDAM
            From ORDTL d Join ORHDR h on d.ONRCU = h.ONRCU
            Where h.CUSNR = :CUSNR and
                  h.ROUTE = :ROUTE and
                  h.STOPX = :STOPX and
                  h.SHPDTISO = :SHPDTISO;

It executed, but the RPG field ORDAM was waAAAyyy off because field

h.ORDAM

from the header file was summed for each record in the detail

file.  I

understand why that is and it makes sense.  I did some googling and

came up

with this, using Distinct:

        Exec SQL
          Select SUM(d.ITNSA),
                 SUM(Distinct(h.ORDAM))
            Into :ITNSA,
                 :ORDAM
            From ORDTL d Join ORHDR h on d.ONRCU = h.ONRCU
            Where h.CUSNR = :CUSNR and
                  h.ROUTE = :ROUTE and
                  h.STOPX = :STOPX and
                  h.SHPDTISO = :SHPDTISO;

That seemed to me like it would have a problem, because 2 selected

order

header records COULD have the same h.ORDAM value.  Won't happen often,

but

it can and does happen sometimes.  If I understand Distinct right, it

would

only include one of them in the Sum function.  So I did some testing

and

found that to be true.

So is there a way to do what I want in one statement?  The flip side

of that

is, if the statement is very complex, I would rather do it in 2

statements

because I believe in KISS.

Thanks.


--
Jeff Crosby
VP Information Systems
UniPro FoodService/Dilgard
P.O. Box 13369
Ft. Wayne, IN 46868-3369
260-422-7531
www.dilgardfoods.com

The opinions expressed are my own and not necessarily the opinion of

my

company.  Unless I say so.
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