Re: Elementary SQL question

[Matthias Oertli <oertlim@xxxxxxxxxxxxxxxx>]

Many thanks for all the replies.

The result I was looking for is one single number (therefore no group
by/order by clauses) which basically is the number of transaction
records divided by the number of customer records giving the average
number of transactions per customers.

The statement sent by Eric DeLong (which put into practice the theory of
Joe Pluta) seems to fit my requirements exactly. I would've never
thought of that trick! Thanks!

Btw, the qmquery manager does not allow nested functions, eg.
avg(count(field)). Is this a limitation that does not exist with the
full blown SQL LPP?

Best regards,
Matthias

>
> How about this (run against order hdr/dtl files):
>
> SELECT (COUNT(DISTINCT char(B.ODORD#)||char(B.ODOELN)) /
>         COUNT(DISTINCT A.OHORD#))
> FROM oepordh a left outer join oepordd b on A.OHORD#=B.ODORD#
>
> The trick is to make sure the numerator (in my case, the order number||Order
> line#) is a unique value. The distinct keyword removes duplicate entries, so
> make sure you account for that.
>

--
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Matthias Oertli   matthias.oertli@mdi.com.au