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j3sting
Regular Visitor

Return Name of User With Highest Number of Clients

Hi,

 

Looking for some help please, im trying to create a measure to retrieve the name of the user with the highest number of clients, my columns are "Keyworker" (this is the user) and "carerid"  (this is the client).

 

I have tried and failed to get this to work, it would be great if someone could point me in the right direction.

 

thanks

 

 

1 ACCEPTED SOLUTION
OwenAuger
Super User I
Super User I

Hi @j3sting

 

You can use a pattern like this (pattern taken from SQLBI - Alternative use of FIRSTNONBLANK and LASTNONBLANK😞

 

TopUser =
FIRSTNONBLANK (
    TOPN (
        1,
        VALUES ( YourTable[Keyworker] ),
        CALCULATE ( DISTINCTCOUNT ( YourTable[carerid] ) )
    ),
    1
)

The FIRSTNONBLANK is just there to break ties.

 

Owen 🙂


Owen Auger

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4 REPLIES 4
v-haibl-msft
Microsoft
Microsoft

@j3sting

 

You can also use the RANKX function to rank the client numbers and get the user name whose rank number is 1.

Assuming we have a table like below.

Return Name of User With Highest Number of Clients_1.jpg

 

We can create two measures with following formulas.

Rank_Client =
RANKX ( ALLSELECTED ( Table1 ), CALCULATE ( SUM ( Table1[carerid] ) ) )
Highest = 
CALCULATE (
    ALLSELECTED ( Table1[Keyworker] ),
    FILTER (
        ADDCOLUMNS ( VALUES ( Table1[Keyworker] ), "RankNum", [Rank_Client] ),
        [RankNum] = 1
    )
)

Return Name of User With Highest Number of Clients_2.jpg

 

Best Regards,

Herbert

thanks Herbert_Liu

 

another great sollution, thanks for the help much appreciated! 🙂

OwenAuger
Super User I
Super User I

Hi @j3sting

 

You can use a pattern like this (pattern taken from SQLBI - Alternative use of FIRSTNONBLANK and LASTNONBLANK😞

 

TopUser =
FIRSTNONBLANK (
    TOPN (
        1,
        VALUES ( YourTable[Keyworker] ),
        CALCULATE ( DISTINCTCOUNT ( YourTable[carerid] ) )
    ),
    1
)

The FIRSTNONBLANK is just there to break ties.

 

Owen 🙂


Owen Auger

Did I answer your question? Mark my post as a solution!

My Blog
Connect on Twitter
Connect on LinkedIn

View solution in original post

Hi OwenAuger 

 

Perfect thanks! 🙂 and thank you for replying 🙂

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