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mwild
Frequent Visitor

String Extract based on criteria

I have a table and one column contains free text.  I'm looking to add a column which will search the free text and extract anything which has a letter followed by 5 numbers.

 

For example - the below would be extracted:


D98765

d12345

D 54821

d 65411

This could be anywhere within the string of text.  Is this possible?

1 ACCEPTED SOLUTION
moizsherwani
Continued Contributor
Continued Contributor

@mwild  ***Made a tweak to the code***

The following works. I will let you tweak the code further if you need to so I am not doing all your homework 😂

 

StringMatch = 
VAR NoBlankFreeText =
    SUBSTITUTE ( 'Sample Data'[FreeText], " ", "", 1 )
VAR UpperD =
    FIND ( "D", NoBlankFreeText, 1, BLANK () )
VAR LowerD =
    FIND ( "d", NoBlankFreeText, 1, BLANK () )
VAR PositionD =
    IF ( ISNUMBER ( UpperD ), UpperD, LowerD)
VAR CheckNumbers =
    ISERROR (
        ISNUMBER ( VALUE ( MID ( NoBlankFreeText, PositionD + 1, 5 ) ) )
    )
RETURN 
    IF (
        NOT ( CheckNumbers ),
        MID ( NoBlankFreeText, PositionD, 6),
        "No Match"
    )

 

 

moizsherwani_0-1632236672765.png

 

 

Thanks,

Moiz
Was I able to answer your question? Mark my post as a solution to help others. Kudos if you liked the solution.

View solution in original post

7 REPLIES 7
mh2587
Solution Sage
Solution Sage

With combination of LEFT or RIGHT function



Muhammad Hasnain

#LetsConnect https://www.linkedin.com/in/hasnain2587/
mh2587
Solution Sage
Solution Sage

Or you can use CONTAINSSTRING function e.g 

MyCalculatedColumn = If(CONTAINSSTRING([TARGETCOLUMN];"searchforthis");TRUE();FALSE())

 



Muhammad Hasnain

#LetsConnect https://www.linkedin.com/in/hasnain2587/
moizsherwani
Continued Contributor
Continued Contributor

I am not sure how this will work when the end goal is "any letter followed by any numbers" when the letter and the numbers could be anything, e.g. it can be "Z 98765" or "b34561" or worse "!@#@!D48202@#!@".

Thanks,

Moiz
Was I able to answer your question? Mark my post as a solution to help others. Kudos if you liked the solution.

The letter will always be "D" or "d"
There may be a space (or may not be) and then 5 numbers between 0 and 9

moizsherwani
Continued Contributor
Continued Contributor

@mwild  ***Made a tweak to the code***

The following works. I will let you tweak the code further if you need to so I am not doing all your homework 😂

 

StringMatch = 
VAR NoBlankFreeText =
    SUBSTITUTE ( 'Sample Data'[FreeText], " ", "", 1 )
VAR UpperD =
    FIND ( "D", NoBlankFreeText, 1, BLANK () )
VAR LowerD =
    FIND ( "d", NoBlankFreeText, 1, BLANK () )
VAR PositionD =
    IF ( ISNUMBER ( UpperD ), UpperD, LowerD)
VAR CheckNumbers =
    ISERROR (
        ISNUMBER ( VALUE ( MID ( NoBlankFreeText, PositionD + 1, 5 ) ) )
    )
RETURN 
    IF (
        NOT ( CheckNumbers ),
        MID ( NoBlankFreeText, PositionD, 6),
        "No Match"
    )

 

 

moizsherwani_0-1632236672765.png

 

 

Thanks,

Moiz
Was I able to answer your question? Mark my post as a solution to help others. Kudos if you liked the solution.
mh2587
Solution Sage
Solution Sage

You can use CONTAIN function



Muhammad Hasnain

#LetsConnect https://www.linkedin.com/in/hasnain2587/
moizsherwani
Continued Contributor
Continued Contributor

You can do this using regular expression (using R or Python) in the query editor, Curbal here explains it really well

 

https://www.youtube.com/watch?v=9E3VsvFAge4

 

Regards,

 

Moiz

If this post helps, please "Accept" it as Solution to help other members find it.

Thanks,

Moiz
Was I able to answer your question? Mark my post as a solution to help others. Kudos if you liked the solution.

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