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viralpatel21's avatar
viralpatel21
Icon for Helper II rankHelper II
5 years ago

Same result in all months error

Hi Team

 

It me again lol. Normally i am good with creating measures but for some reason i am not getting a monthly breakdown in my table. I am getting the same total for all months. it even create a total for a blank month? So what the measure is doing its counting the number of members from the accounts table but its filtering if there been more that 1 transaction from the BalanceAdjustment table.

VolumeofID =
CALCULATE(
DISTINCTCOUNT(BalanceAdjustments[TransactionId]),
FILTER(
BalanceAdjustments,
BalanceAdjustments[Load_Type] = "Bank Transfer" ||
BalanceAdjustments[Load_Type] = "Digital Asset" ||
BalanceAdjustments[Load_Type] = "TopUP"),
USERELATIONSHIP(BalanceAdjustments[Date],FeeDate[Date])
)

 

 

Relationships:

FeeDate[Date] 1 to many Aufsaaccounts[createdon]

FeeDate[Date] 1 to mant BalanceAdjustments[Date]

BalanceAdjustment[AccountNumber] Many to many a au_fsaccountses[fsaacountreference]

 

Im not sure what im doing wrong. I checked all the dates format and it is correct?

 

thanks

viral

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3 Replies

    • viralpatel21's avatar
      viralpatel21
      Icon for Helper II rankHelper II
      5 years ago

      hi amitchandak 

       

      thanks for the reply. Im still getting confused on how to do that?

      But if i didnt remove the M-M relationship because there are more tables that factors into the data model... how would i fix the current problem?

       

      thanks

      viral

  • Anonymous's avatar
    Anonymous
    Not applicable
    5 years ago

    Hi viralpatel21 

    I think it may be the many to many relationship that caused the measure you created to return the same result .

    Could you provide some data or sample ? According to your diagram, I have no way to determine the specific reason .

     

    Best Regards

    Community Support Team _ Ailsa Tao

     

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.