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smilingjohn
Specialist
Specialist

List box

Hi Expertise

I am using the below expression  in my pivot table to display the weekly data . and this is a running total  And I am using this part in my expression (Sum({1<Year *= {">2015"}>}0), 0) to populate the missing weeks so that I will be dipsplaying all the weeks  and the result is accurate

 

 

The only problem I am facing is that  the data is is diplaying from 2016 onwards , and if I select Year=2019 it doesn’t  change.

 

The table should show the weekly data according to my Year list box selection .Can you please help me on this .

I got have fixed this using possible values . 

 

Thanks in Advance

4 Replies
jonathandienst
Partner - Champion III
Partner - Champion III

The '1' ignores all selections, so the intersection is between everything and 2015, and your selection will not be considered. What goes wrong if you remove the '1' (or replace with '$')?

Logic will get you from a to b. Imagination will take you everywhere. - A Einstein
smilingjohn
Specialist
Specialist
Author

Hi Jonty Thanks for the reply , 

If i replace 1 with $ or nothing then upon selecting year 2019 , it shows me only the weekly dates where there is data . 

but i need to see all the weekly dates within it thats the reason i  used 1. 

I want to display all the week with list box selection . Please loo at the attachment

 

Thanks 

 

 

smilingjohn
Specialist
Specialist
Author

Any Help on this ?

Brett_Bleess
Former Employee
Former Employee

John, have a look at the following Design Blog post and I am including the base URL after that in case you want to do some further searching:

https://community.qlik.com/t5/Qlik-Design-Blog/A-Primer-on-Set-Analysis/ba-p/1468344

https://community.qlik.com/t5/Qlik-Design-Blog/bg-p/qlik-design-blog

You will need to use an expression in the set analysis instead of the year value in this case that ties to the list box etc...  

https://help.qlik.com/en-US/qlikview/April2019/Subsystems/Client/Content/QV_QlikView/ChartFunctions/...

Regards,
Brett

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