Query question - can I return values for records not found?

My database (MS SQL2K) is used primarily for the entry, tracking and reporting of (software) QA test results. This query is used to poulate a test matrix page - columns represent supported platforms, rows represent test cases. Each cell is a link to the data entry page where the test case and platform are prepopulated. (Also prepopulated - from a cookie - are test date, build, language, locale and QA engineer, so that only pass or fail+severity and defect # need to be entered.) Submit returns the user to the updated matrix page, which is populated by the query in question, here...

For each combination of test case and supported platform, I need one record to be returned - even if there's no data in the TestResults table for some combinations. If there is data I want the query to return TestCase, Platform, TestDate, Build, QAEngineer and Result for the most recent test result (using MAX statements for Build and TestDate) - BUT, for combinations of test case/platform that have not yet been tested, is there some way to tell the query to return a record with "UNTESTED" (or even NULL) values for each (or even the first) field in the record?

How is the data stored?
What's your table structure?

You can achieve what you mentioned above if you do something like this:
SELECT {fields} FROM Software s RIGHT JOIN Test t {etc...}

RIGHT JOIN causes the query to return all records from the first table, even if there isn't a matching record on the second table.

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Diego - Web-Developer & London SEO Expert
jQuery Plugins: Multiple File Upload, Star Rating, FCKEditor | NEW: XML to JSON

Use a default.

a default specifies a value for a null field in a new record...
it's got nothing to do with selects.
no records = no fields (using defaults or not)

__________________
Diego - Web-Developer & London SEO Expert
jQuery Plugins: Multiple File Upload, Star Rating, FCKEditor | NEW: XML to JSON

Turns out what I needed was a cross join combined with a left outer join. Pretty complicated, but works beautifully in the end.

Sounds pretty inefficient for what you wanted to do, if you find it's slow or needs tweaking, post the query here.