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Need to find ID's that are common in both sets. Below Expression is not working!
=count({<[ID] = P({<Month_Year = [set1]::Month_Year>})*P({<Month_Year = [set2]::Month_Year>})>}ID)
Did you try like this?
=count({<[ID] = P({[set1]<Month_Year = $::Month_Year>}[ID]) * P({[set2]<Month_Year = $::Month_Year>} [ID]) >} ID)
Or this
=Count({<[ID] = P({1<Month_Year = [set1]::Month_Year>})*P({1<Month_Year = [set2]::Month_Year>})>}ID)
I am having different answers using above
=Count({<[ID] = P({1<Month_Year = [set1]::Month_Year>})*P({1<Month_Year = [set2]::Month_Year>})>}ID)
=count({<[ID] = P({[set1]<Month_Year = $::Month_Year>}[ID]) * P({[set2]<Month_Year = $::Month_Year>} [ID]) >} ID)
and when I am seeing individual sets counts
Count({$*set1}ID) is not giving me same result in Count({<[ID] = P({1<Month_Year = [set1]::Month_Year>})ID)
How do I get common ID's when two months are selected in alternative sets of Month_Year filter.
If you want to find IDs that are common in both sets, this should be enough:
Need to find ID's that are common in both sets. Below Expression is not working!
=count({<[ID] = P({set1}) * P({set2})>} ID)
If your ID field is a key field, don't use a plain count(). Add a DISTINCT qualifier or count a non-key field.
=count({<[ID] = P({set1}) * P({set2})>} DISTINCT ID)
Also note that your current / inherited selections in other fields might interfere with your set modifier.