Forum Discussion

guenfood's avatar
guenfood
New Member
4 years ago
Solved

Rank by index

Hi,

 

I've got a table with an index column.

To display many text informations in cards, i create measures which could display values according to ranking position.

But this ranking position must be based on index number.

I tried with this measure but i've got a message error "Unable to determine the value of "index". Column does not exist or there is no current row for this column

 

GlobalRANK = RANKX(BDD, [Index])

 

 I had search on the forum and internet but did'nt fin anything to solve my problem.

Is anyone could help me ?

Need Help
  • v-henryk-mstf's avatar
    v-henryk-mstf
    4 years ago

    Hi guenfood ,

     

    According to your description, I did the test reference as follows:

    Column = 
"Rank"
    & RANKX (
        FILTER (
            ALL ( 'Table (2)' ),
            'Table (2)'[Index] = EARLIER ( 'Table (2)'[Index] )
        ),
        [Value],
        ,
        ASC,
        DENSE
    )


    If I have misunderstood your requirements, please point out and provide some test data and screenshots of the desired results so that I can fix them as soon as possible. Looking forward to your reply.


    Best Regards,
    Henry


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

     

3 Replies

  • amitchandak's avatar
    amitchandak
    Super User
    4 years ago

    guenfood , Seem like you are trying to create a measure, for the measure you need a measure


    GlobalRANK = RANKX(BDD, calculate(count([Index])))


    Column Rank can be done like

    GlobalRANK = RANKX(BDD, [Index])

    • guenfood's avatar
      guenfood
      New Member
      4 years ago

      Thanks for your answer.

      I tried with this formula but i don't understand.

      It seems to return only one Rank value.

      But my index have several lines (actually, 5 lines for a single days), so i must have 5 differents values for RANK.
      RANK 1

      RANK 2

      RANK 3

      RANK 4 

      RANK 5

  • v-henryk-mstf's avatar
    v-henryk-mstf
    Community Support
    4 years ago

    Hi guenfood ,

     

    According to your description, I did the test reference as follows:

    Column = 
"Rank"
    & RANKX (
        FILTER (
            ALL ( 'Table (2)' ),
            'Table (2)'[Index] = EARLIER ( 'Table (2)'[Index] )
        ),
        [Value],
        ,
        ASC,
        DENSE
    )


    If I have misunderstood your requirements, please point out and provide some test data and screenshots of the desired results so that I can fix them as soon as possible. Looking forward to your reply.


    Best Regards,
    Henry


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