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daviss62
Frequent Visitor

Switch True conditions for values in the same column but different rows

Hello.

 

I am trying to use Switch True that would return a specific value is a particular customer has a certain combination of values (from the same column). However, a customer can have more than one row of data--meaning one value being on one row, and the other value on another row. I want to be able to determine whether or not each customer has the specified Code values from that column (though each Code value is on a different row for the same customer), and then return the appropriate resulting "Code Combination" value.

 

When I try the above approach with my actual data table, the new "Code Combination" column returns blanks. Any assistance is greatly appreciated.

 

For example:

Customer ID     Code        

01737298         WBSA

01737298         WMSA

01568489         WBSM

 

Code Combination =

SWITCH(
     TRUE(),

     'Table'[Code] = "WBSA" && 'Table[Code] = "WMSA", "BS to MS")

3 REPLIES 3
v-diye-msft
Community Support
Community Support

Hi @daviss62 

 

Not sure if you'd like to get below result:

Measure 2 = var a = CALCULATE(MAX('Table (3)'[Customer ID]),FILTER('Table (3)',[Code]="WBSA"))
var b = CALCULATE(MAX('Table (3)'[Customer ID]),FILTER('Table (3)',[Code]="WMSA")) 
Return
IF(a=b,BLANK(),"BS to MS")

16.PNG

Community Support Team _ Dina Ye
If this post helps, then please consider Accept it as the solution to help the other members find it more
quickly.

Hi @v-diye-msft ,

 

That is close to what I want. However, when I plug in my columns/fields into a new calculated column, I get the following "circular dependency" error message.

 

2019-09-17_15h09_42.png

Hi @daviss62 

 

 I am aware of a circular dependency "issue" when you write multiple calc columns, and their incrementally refer to all columns in a table. Did you use calculated com which should be measure instead?

 

Community Support Team _ Dina Ye
If this post helps, then please consider Accept it as the solution to help the other members find it more
quickly.

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