Forum Discussion
KPI value for multi-select months in slicer
Hi Everyone,
I wanted to visualize current month data with respect to last month to compare the performance of a metric and I am using KPI card for that. This is working fine when I am using current month and previous month (target). But when in slicer I am selecting multi-select (Ctrl + Click), it is displaying only the latest updated data and not the sum of all months.
For ex - For June 2023, the value is 100 and for May 2023 it is 50. When clicking both the months it should display 150 as answer but it is giving me 100 (the latest one which is selected/clicked in slicer).
Is there any way I can solve this problem?
- Anonymous3 years ago
Hi hvshalihotri ,
You can try the following dax:
Measure = var _selectyear=SELECTEDVALUE('Table'[Date].[Year]) var _selectmonth=SELECTCOLUMNS('Table',"1",'Table'[Date].[Month]) return SUMX( FILTER(ALL('Table'), YEAR('Table'[Date])=_selectyear&&'Table'[Date].[Month] in _selectmonth),[Value])Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 Replies
- lbendlin
Super User
Please provide sample data (with sensitive information removed) that covers your issue or question completely, in a usable format (not as a screenshot).
https://community.fabric.microsoft.com/t5/Community-Blog/How-to-provide-sample-data-in-the-Power-BI-Forum/ba-p/963216
Please show the expected outcome based on the sample data you provided.
https://community.fabric.microsoft.com/t5/Desktop/How-to-Get-Your-Question-Answered-Quickly/m-p/1447523 - AnonymousNot applicable
Hi hvshalihotri ,
You can try the following dax:
Measure = var _selectyear=SELECTEDVALUE('Table'[Date].[Year]) var _selectmonth=SELECTCOLUMNS('Table',"1",'Table'[Date].[Month]) return SUMX( FILTER(ALL('Table'), YEAR('Table'[Date])=_selectyear&&'Table'[Date].[Month] in _selectmonth),[Value])Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.