I figured it out, and it was much easier than I thought 🙂

=iif(left(Parameters!Search.Value,6) = "Amount","'$'#,0","#,0")

This works great, but is only accounting for those with FieldID =1. Would you just double the code and change your WHERE statement to FieldID = 2?

Sean Lange (2/6/2012)

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Pretty sure you can handle the counts now.

As to getting the resultset of all possibilities...

What do you want that to look like? I know it was just assigned...

Sean Lange (2/6/2012)

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mmiller 85218 (2/6/2012)

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Well, they want a count of all total possibilities...with that, i can apply...

Well, they want a count of all total possibilities...with that, i can apply the factoral mathematics and...

J Livingston SQL (2/6/2012)

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mmiller 85218 (2/6/2012)

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Sean Lange (2/6/2012)

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J Livingston SQL (2/6/2012)

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2 type of House

4 type of Colour

3 type of Size

5 type of Flavour

2*4*3*5 = 120 combinations

though how they...

Sean Lange (2/6/2012)

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J Livingston SQL (2/6/2012)

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2 type of House

4 type of Colour

3 type of Size

5 type of Flavour

2*4*3*5 = 120 combinations

though how they are related beats me 😀

edit...

LOL They aren't, its just replica data of actual data I'm working with 🙂

I have read up on the combinations code, and will view the links you sent. I know the request is odd, I have told the requestor that there are...

I just know by the data but for this sample table, I am assuming that CatID 1 and CatID 3 can only have one option, the other 2 can have...

I need the actual possiblities because I will need to tie the results to actual programs we run and the combinations of all of them.

The results are based on the FieldId. So the first example 1/3/7/10 would be

CatId 1 FieldId 1

CatId 2 FieldId 3

CatId 3 FieldId 7

CatId 4 FieldId 10

FieldID could be replacedwith...

They aren't represented in the database, I would just code for it if possible.

I need results as they show in my first post. So the actual combination of the different possiblities.