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Microsoft Power Up Program Community
Answered

Current Location Power FX formula

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Posted on by 12
I wrote this formula for the current location column of the animals table and I get the error "result type void is not supported." How can I fix this?
If(
    IsBlank('Adopter Name'),
    If(
        IsBlank('Foster Claimer'),
        "Initial Shelter",
        'Foster Claimer'
    ),
    'Adopter Name'
)
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I have the same question (0)
  • w.p Profile Picture
    8,339 Super User 2025 Season 2 on at
    I think it should be based on shelter status field.
     
    This is the sample data.
  • CU13011934-0 Profile Picture
    20 on at
    Hi 
     
    Did you resolve this query? As I  am struggling with the same issue?
     
     
  • HK-06012306-0 Profile Picture
    12 on at
    I tried writing a formula with the shelter status column, and I am still running into the same result type void is not supported error. Here is my formula based on shelter status: 
     
    Any suggestions on how to remedy this error? Thanks!
    If(
        'Shelter Status' = "Ready for Adoption" Or 'Shelter Status' = "Claimed for Adoption" Or 'Shelter Status' = "In Foster Home", 
        'Foster Claimer', 
        If(
            'Shelter Status' = "In Shelter", 
            'Initial Shelter', 
            "N/A"
        )
    )
    
  • w.p Profile Picture
    8,339 Super User 2025 Season 2 on at
     
  • Suggested answer
    w.p Profile Picture
    8,339 Super User 2025 Season 2 on at
  • Suggested answer
    timl Profile Picture
    36,391 Super User 2025 Season 2 on at
     
    In answer to your initial question, the most likely issue is that 'Foster Claimer' is setup as a lookup, therefore you would need to specify a field from the 'Foster Claimer' record such as 'Foster Name'.
     
    If(
        IsBlank('Adopter Name'),
        If(IsBlank('Foster Claimer'),"Init", 'Foster Claimer'.'Foster Name'),
        'Adopter Name'
    )
    ​I tested this and don't get the error.
     
     
     
    @CU13011934-0 -  since you're also experiencing this problem, can you confirm if the above resolves the issue for you?
  • Verified answer
    timl Profile Picture
    36,391 Super User 2025 Season 2 on at
     
    In answer to your second question about the shelter status, there are 2 things to highlight about the syntax you used. 
     
    As @W.P suggests below, because  'Shelter Status' is a choice column, the correct way to carry out this type of conditional test 'Shelter Status' = "Ready for Adoption" is to reference the enumeration.
     
    For a global choice set, it would look like this - 'Shelter Status' = [@ShelterStatus]'Ready for Adoption'
    For a local choice set, it would look like this ' - 'Shelter Status' = 'Shelter Status (Animals)'.'Ready for Adoption'
     
    The second point is the same as what I mentioned below for 'foster claimer'. If 'Initial Shelter' is a lookup, you would need to specify one of the fields such as 'Initial Shelter'.'Shelter Name'.
     
    Therefore, the working version of the code you posted below would look like this:
    If(
        'Shelter Status' = 'Shelter Status (Animals)'.'Ready for Adoption' Or 'Shelter Status' = 'Shelter Status (Animals)'.'Claimed for Adoption' Or 'Shelter Status' = 'Shelter Status (Animals)'.'In Foster Home', 
        'Foster Claimer'.'Foster Name', 
        If(
            'Shelter Status' = 'Shelter Status (Animals)'.'In Shelter', 
            'Initial Shelter'.'Shelter Name', 
            "N/A"
        )
    )
     
  • w.p Profile Picture
    8,339 Super User 2025 Season 2 on at
    My BI version.
  • timl Profile Picture
    36,391 Super User 2025 Season 2 on at
    Nice work in the PowerBI @W.P - that's looking good!

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