× The internal search function is temporarily non-functional. The current search engine is no longer viable and we are researching alternatives.
As a stop gap measure, we are using Google's custom search engine service.
If you know of an easy to use, open source, search engine ... please contact support@midrange.com.



Add a row with a NULL value for ORDER# to see the difference. See the nuances of COUNT here, for row versus value count and the NULL value.
http://publib.boulder.ibm.com/infocenter/iseries/v5r4/topic/db2/rbafzmstcolfunc.htm

Regards, Chuck

rob@xxxxxxxxx wrote:
I do not think your "count" implementation works. COUNT seems to
ignore what's in the parenthesis.

select co, cust, PO#, order#, pick#
from qtemp.craig 1....+....2....+....3....+....4....+....5....+....6....+.
CO CUST PO# ORDER# PICK#
100 101 a1 1 1
100 101 a1 1 2
100 101 a1 1 3
100 101 a1 2 4
100 101 a1 2 5
End of data ********
select co, cust, PO#, count(order#) AS NbrOrders, count(pick#) as
nbrPicks from qtemp.craig group by co, cust, po# 1....+....2....+....3....+....4....+....5....+....6....+..
CO CUST PO# NBRORDERS NBRPICKS 100 101 a1 5 5 End of data ********
Perhaps with some complicated mess involving DISTINCT or some such thing...

Rob Berendt

As an Amazon Associate we earn from qualifying purchases.

This thread ...


Follow On AppleNews
Return to Archive home page | Return to MIDRANGE.COM home page

This mailing list archive is Copyright 1997-2024 by midrange.com and David Gibbs as a compilation work. Use of the archive is restricted to research of a business or technical nature. Any other uses are prohibited. Full details are available on our policy page. If you have questions about this, please contact [javascript protected email address].

Operating expenses for this site are earned using the Amazon Associate program and Google Adsense.