Adding year to measure

[wnicholl]

Hello,

I need to add a year to the below formula. I tried to add; DATESBETWEEN('Calendar'[Date], DATE(2017,01,01), DATE(2017,31,12) to the formula however when I add as a value it can't display the visual. Is there something that I'm missing or is there another option to add dates?    

Count of Active Policy  =

DIVIDE(

    DISTINCTCOUNT('Policy'[Active Policy - Accnt ID]),

    DISTINCTCOUNT('Account'[Full_Account_ID__c])

)

[Nathaniel_C · Accepted solution]

Hi @wnicholl ,
Please try this:

Your tables will need to be linked in the data model usually Calendar to Policy and Calendar to Account.

 

Count of Active Policy  =

CALCULATE(
DIVIDE(
    DISTINCTCOUNT('Policy'[Active Policy - Accnt ID]),
    DISTINCTCOUNT('Account'[Full_Account_ID__c])
)
FILTER(CALENDAR, CALENDAR[DATE]>= DATE(2017,01,01) && CALENDAR[DATE]<= DATE(2017,31,12)))

Let me know if you have any questions.

If this solves your issues, please mark it as the solution, so that others can find it easily. Kudos 👍are nice too.
Nathaniel

[Ashish_Mathur · Accepted solution]

Hi,

Does this measure work?

Count of Active Policy  = calculate(DIVIDE(DISTINCTCOUNT('Policy'[Active Policy - Accnt ID]),DISTINCTCOUNT('Account'[Full_Account_ID__c])),DATESBETWEEN('Calendar'[Date], DATE(2017,01,01), DATE(2017,31,12))

[wnicholl]

Thanks for your reply. You measure worked for me. 

[Nathaniel_C · Accepted solution]

Hi @wnicholl ,
Please try this:

Your tables will need to be linked in the data model usually Calendar to Policy and Calendar to Account.

 

Count of Active Policy  =

CALCULATE(
DIVIDE(
    DISTINCTCOUNT('Policy'[Active Policy - Accnt ID]),
    DISTINCTCOUNT('Account'[Full_Account_ID__c])
)
FILTER(CALENDAR, CALENDAR[DATE]>= DATE(2017,01,01) && CALENDAR[DATE]<= DATE(2017,31,12)))

Let me know if you have any questions.

If this solves your issues, please mark it as the solution, so that others can find it easily. Kudos 👍are nice too.
Nathaniel

[wnicholl]

Thanks for your reply. You measure worked for me. 

[Nathaniel_C]

Hi @wnicholl ,
You are welcome!

Nathaniel