Forum Discussion
Select value from slicer in another table and use in filter
Hi,
I'd like to ask for help please.
I have a table table1 which contains "Sprint name" column and is used in slicer.
I have another table table2 which is summarized/filter from table1. I want to use the selectedvalue from slicer to be able to filter my table2 but with a different value:
filter(table2, table2[index] <= (select index from table2 where [table2]sprintname = selectedvalue(table1[sprintname]))
I'm not certain how to do the one in bold. When i create a var as part of table2 query:
var sprintName = selectedvalue(table1[sprintname]) <- this one doesnt seem to pick up any value
but when i create a measure under table2, i am able to get the selectedvalue but i'm not able to access it as filter in the table2 query (where measure was created).
Thanks so much!
3 Replies
- AnonymousNot applicable
Have you tried just =, instead of <=? The SelectedValue is a text value, right?
--Nate
- AnonymousNot applicable
Hi! Unfortnulately, no that doesn't work.
I wanted to filter may table2 based on the selected value from slicer in table1
If I hard code sprint name below it works:
Table2 =VAR Result = SUMMARIZE(FILTER('Table1', some filters), 'Table1'[Sprint Name], 'Table1'[Iteration Start Date])VAR FilteredResult = FILTER(Result, 'Table1'[Sprint Name] <= "Sprint 17")Return FilteredResultIf i don't hard code, and get value from measure or selectedValue from table1, it doesn't work.How can I get selectedvalue from slicer from table1?Thanks so much!- v-yingjl
Community Support
Hi Anonymous ,
Currently it is not supported to pass selected slicer values to a calculated table in Power BI, use selectedvalue() would return all blank value.
If write like calculate(max('Table'[Name]),allselected('Table')), it would return the max name value from the table.
You have need to make hard code for the calculated table to filter it.
Best Regards,
Community Support Team _ Yingjie Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.