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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 |
Solved! Go to Solution.
@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
@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
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