<> represents the Null fields of the left join with query1.
This is untested but try it --
Query2:
SELECT qryCategories.CatID, qryCategories.CatDesc, Nz(Query1.Step1Decision,
"None") AS Decision
FROM qryCategories LEFT JOIN Query1 ON qryCategories.CatID = Query1.CatID;
Query3:
TRANSFORM Count(Query2.Step1Decision) AS [Count]
SELECT Query2.CatID, Query2.CatDesc, Count(Query2.CatID) AS Total
FROM Query2
GROUP BY Query2.CatID, Query2.CatDesc
PIVOT Query2.Decision;
--
Build a little, test a little.
"ripper" wrote:
> I have Query1 that returns 2 fields:
> CatID
> Step1Decision
>
> CatID can be 01 thru 29
> Step1Decision can be N,X,P,D,R
>
> Query1:
> SELECT qryGrievances.CatID, qryGrievances.Step1Decision
> FROM qryGrievances, qryStartEnd
>
> Query2 uses Query1 joined to qryCategories to return all the categories and
> their description plus those in Query1
>
> Query2:
> SELECT qryCategories.CatID, qryCategories.CatDesc, Query1.Step1Decision
> FROM qryCategories LEFT JOIN Query1 ON qryCategories.CatID = Query1.CatID
>
> Query3 uses Query2 in a crosstab so the Step1Decision (N,X,P,D,R) become
> column headings and categories become rows, and adds a Total field:
>
> Query3:
> TRANSFORM Count(Query2.Step1Decision) AS [Count]
> SELECT Query2.CatID, Query2.CatDesc, Count(Query2.CatID) AS Total
> FROM Query2
> GROUP BY Query2.CatID, Query2.CatDesc
> PIVOT Query2.Step1Decision
>
> Result looks like this:
>
> CatID CatDesc Total <> D N R X
> 01 Accounting 7 6 1
> 02 Assignments 24 5 12 7
> 03 Conditions 1 0
> 04 Disciplinary 2 2
>
> My questions are: how is the <> field generated? I can't refer to it in
> code, which is a problem. Also, why does the total column show correct
> totals for the D,N,R,X values added across, but if there are no records
> returned for a category, it still shows 1? Also, if no records exist for one
> of the decision values (P in the case above), that field is not returned by
> the query at all. How can I create a report based on a query that may or may
> not return some of the fields?
>
> Thanks to all who can help!
> Ripper
>
>
> .
>
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