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Set Analysis - Basic question about intersection operator (*)

I've read up on a few threads about this but haven't found the specific answer that I'm looking for. Basically, I'm trying this expression for a straight table:

= count({<[Gifts Type Code] = {c}> * <[Gifts Honor Marker] = {B}>} [Gifts ID])

and it yields 0. However if I use this

= count({<[Gifts Type Code] = {c}, [Gifts Honor Marker] = {B}>} [Gifts ID])

It works. My question is why? It seems like the two statements should yield identical results

Obviously for this specific case I can use the second statement, but I use the intersection operator (*) in several other places so I want to make sure I know what it actually does.

Thanks

1 Solution

Accepted Solutions
pover
Luminary Alumni
Luminary Alumni

The syntax looks fine, but I've found a difference in how each responds to user selections because if the user makes a selection of a [Gifts Hone Marker] that isn't B than the 2 sets don't intersect so

= count({<[Gifts Type Code] = {c}, [Gifts Honor Marker] = {B}>} [Gifts ID])

would be more like

= count({<[Gifts Honor Marker]=, [Gifts Type Code] = {c}> * <[Gifts Honor Marker] = {B},[Gifts Type Code]=>} [Gifts ID])

Regards.

View solution in original post

2 Replies
pover
Luminary Alumni
Luminary Alumni

The syntax looks fine, but I've found a difference in how each responds to user selections because if the user makes a selection of a [Gifts Hone Marker] that isn't B than the 2 sets don't intersect so

= count({<[Gifts Type Code] = {c}, [Gifts Honor Marker] = {B}>} [Gifts ID])

would be more like

= count({<[Gifts Honor Marker]=, [Gifts Type Code] = {c}> * <[Gifts Honor Marker] = {B},[Gifts Type Code]=>} [Gifts ID])

Regards.

Not applicable
Author

Thanks a lot, that's very helpful.