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blestone111
Helper I
Helper I

Yes of No based on Reference Number in a column

Hi Experts

 

Whats the best method to tackle the following - See sample data below. If the column below has BOTH number and Text then Y other wish N

 

Reference
1245T
123456t
who knows
12345t
93837474H
12345t
12345t
1
567
0878gkj
086425
98765G
87624nk
87624nk
87624nk
87624nk
92837h
98765
12345t
1234
12345678h
102345t
notrace
123456y
12345t
1 ACCEPTED SOLUTION
tharunkumarRTK
Super User
Super User

@blestone111 
There are many ways to accompilsh, here is one way

you can paste the below formula as a conditional column 

if ( List.Count(Splitter.SplitTextByCharacterTransition({"A".."Z", "a".."z"},{"0".."9"})([Reference]))) > 1 or (
 List.Count(Splitter.SplitTextByCharacterTransition({"0".."9"}, {"A".."Z", "a".."z"})([Reference]))   
) > 1 then "Yes" else "No"


 


If the post helps please give a thumbs up


If it solves your issue, please accept it as the solution to help the other members find it more quickly.


Tharun

 

View solution in original post

2 REPLIES 2
tharunkumarRTK
Super User
Super User

@blestone111 
There are many ways to accompilsh, here is one way

you can paste the below formula as a conditional column 

if ( List.Count(Splitter.SplitTextByCharacterTransition({"A".."Z", "a".."z"},{"0".."9"})([Reference]))) > 1 or (
 List.Count(Splitter.SplitTextByCharacterTransition({"0".."9"}, {"A".."Z", "a".."z"})([Reference]))   
) > 1 then "Yes" else "No"


 


If the post helps please give a thumbs up


If it solves your issue, please accept it as the solution to help the other members find it more quickly.


Tharun

 

Hello  tharunkumarRTK,
 if i want to create a same thing but using DAX the which dax used.
i try containstring , find search but not give the appropiate result.

Thanks in Andavance

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