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Anonymous's avatar
Anonymous
Not applicable
3 years ago
Solved

Power Bi - Measured column

I have 3 tables, each one is a parent of other.

 

Initiative-->Epic -->Feature

 

All these tables have a column named "Status".

The column "status" has 2 values "Done" or "Inprogress"

 

I need a card which shows the completion status in % of a epic.

 

For example, a epic has 4 features, out of these, 2's status is "Done" and remaining 2's status is "Inprogress".

 

So, in card it should show as 50% of a selected epic

 

Kindly help me here,

Thanks in advance

  • Anonymous's avatar
    Anonymous
    3 years ago

    Hi Anonymous ,

     

    Sorry, still not very clear. Could you go into more detail and provide some simple sample data and the corresponding expected results? Thank you.

    Which state do you want to calculate the percentage?

    percentage = var _totalgroupbyepic=COUNTROWS(FILTER(ALLSELECTED('Table'),[epic]=MAX('Table'[epic])))
    var _count=COUNTROWS(FILTER(ALLSELECTED('Table'),[status]="Done"&&[epic]=MAX('Table'[epic])))
    return DIVIDE(_count,_totalgroupbyepic)

     

     

     

    Best Regards,

    Stephen Tao

     

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

  • Anonymous's avatar
    Anonymous
    3 years ago

    Hi Anonymous ,

    Your solution is right. 

    Thank you for your reply : )

     

2 Replies

  • Anonymous's avatar
    Anonymous
    Not applicable

    Hi Anonymous ,

     

    Sorry, still not very clear. Could you go into more detail and provide some simple sample data and the corresponding expected results? Thank you.

    Which state do you want to calculate the percentage?

    percentage = var _totalgroupbyepic=COUNTROWS(FILTER(ALLSELECTED('Table'),[epic]=MAX('Table'[epic])))
    var _count=COUNTROWS(FILTER(ALLSELECTED('Table'),[status]="Done"&&[epic]=MAX('Table'[epic])))
    return DIVIDE(_count,_totalgroupbyepic)

     

     

     

    Best Regards,

    Stephen Tao

     

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

    • Anonymous's avatar
      Anonymous
      Not applicable

      Hi Anonymous ,

      Your solution is right. 

      Thank you for your reply : )