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Date filter hierarchy is showing all months even though the value for that month is not present

Hello All,

 

I have a date column in my dataset and using that date column-date hierarchy to filter table values.

 

Date hierarchy silcer is showing all months and days even though the values for those months are not avaiable.

 

Date column:

Date column.png

 

Date Filter

Date Filter.png

 

We would like to show only 2022 year, October Month, 19 day in above filter example.

 

Please let me know how to do that.

 

Thank you!

 

Regards,

Ashwini

 

 

1 ACCEPTED SOLUTION

Below measure worked for me. Thank you for your time @amitchandak 

 

Created a measure like below, and applied it as a visual level filter on the slicer, filtered to 1.

 

SlicerFilter = INT ( NOT ISEMPTY ( FactTable ) )

View solution in original post

3 REPLIES 3

Below measure worked for me. Thank you for your time @amitchandak 

 

Created a measure like below, and applied it as a visual level filter on the slicer, filtered to 1.

 

SlicerFilter = INT ( NOT ISEMPTY ( FactTable ) )

Tried using Date table date column too.. It is also showing all months and days..

amitchandak
Super User
Super User

@ashuaswinireddy , better to create a date table with year, month, day etc., and use that in the slicer 

 

Calendar = Addcolumns(calendar(date(2020,01,01), date(2021,12,31) ), "Month no" , month([date])
, "Year", year([date])
, "Month Year", format([date],"mmm-yyyy")
, "Month year sort", year([date])*100 + month([date])
, "Qtr Year", format([date],"yyyy-\QQ")
, "Qtr", quarter([date])
, "Month",FORMAT([Date],"mmmm")
, "Month sort", month([DAte])
, "Is Today" ,if([Date]=TODAY(),"Today",[Date]&"")
,"Day of Year" , datediff(date(year([DAte]),1,1), [Date], day)+1
, "Month Type", Switch( True(),
eomonth([Date],0) = eomonth(Today(),-1),"Last Month" ,
eomonth([Date],0)= eomonth(Today(),0),"This Month" ,
Format([Date],"MMM-YYYY") )
,"Year Type" , Switch( True(),
year([Date])= year(Today()),"This Year" ,
year([Date])= year(Today())-1,"Last Year" ,
Format([Date],"YYYY")
)
)

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