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fredericgarnier69
Contributor II
Contributor II

Use a variable with Dimension of Pivot Table

Hello

I need some help 😉

I have dimension AA with values 'one' and 'two'

I have another dimension BB with values 'red' and 'green'

I have several  variables:

one_red = 1,  one_green=2, two_red=3, two_green=4

 

how can I build my formula so that when looping through the table, each cell in the table is the right combination?

 

Pivot table: 

               red  |  green

one          1          2

two           3         4

 

I tried : $(=Only(AA)) & '_' & $(=Only(BB))   right but not evaluated as a varible 😞

Any help will be welcome... Thank's in advance

 

Labels (3)
6 Replies
Rohan
Specialist
Specialist

Hi,

Try :

Pick(AA,

pick(wildmatch(BB,'Red','Green'),$(vone_red),$(vone_green)),

pick(wildmatch(BB,'Red','Green'),$(vtwo_red),$(vtwo_green))

)

Regards,

Rohan.

fredericgarnier69
Contributor II
Contributor II
Author

Thank's Rohan

This indeed works, but I want to avoid using the pick and match functions because it is very consuming and in fact, the pivot table is very slow to display.

The solution I am considering must be based on the constitution of a variable...

 

Thank's a lot

BR

Chanty4u
MVP
MVP

Try this 

$(=Only(AA) & '_' & Only(BB))

 

fredericgarnier69
Contributor II
Contributor II
Author

Unfortunetly: Error in expression 😞

 

Rohan
Specialist
Specialist

Try :

$($(=maxstring(AA)&_&maxstring(BB)))

 

Regards,

Rohan.

fredericgarnier69
Contributor II
Contributor II
Author

Unfortunately, this is not interpreted as the content of the variable 😞