Forum Discussion

yasbos's avatar
yasbos
Resolver II
2 years ago
Solved

conditional all()??

Hi. I have a calculate function. I need to decide on the fly whether it uses all() (based on user input in a checkbox) as one of its arguments. Is this at all possible? Calculate( something, if(check...
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  • Anonymous's avatar
    Anonymous
    2 years ago

    Hi yasbos ,

     

    Maybe you can try formula like below:

    ConditionalTotalSales = 
IF(
    [SelectedOption] = "Include All",
    CALCULATE(
        [TotalSales],
        ALL('Calendar')
    ),
    CALCULATE(
        [TotalSales]
    )
)
    SelectedOption = SELECTEDVALUE(Checkbox[Option], "Exclude All")

    TotalSales = SUM(Sales[Sales])

     

    Best Regards,
    Adamk Kong

     

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

  • yasbos's avatar
    yasbos
    2 years ago

    Thanks so much, lbendlin . I forgot to mention that I did know this solution, but I was hoping for a more localized decision because the calculate function that I have is much larger than just one line. I was hoping to modularize the code. Thanks so much for your help, nonetheless.

Anonymous's avatar
Anonymous
Not applicable
2 years ago

Hi yasbos ,

 

Maybe you can try formula like below:

ConditionalTotalSales = 
IF(
    [SelectedOption] = "Include All",
    CALCULATE(
        [TotalSales],
        ALL('Calendar')
    ),
    CALCULATE(
        [TotalSales]
    )
)
SelectedOption = SELECTEDVALUE(Checkbox[Option], "Exclude All")

TotalSales = SUM(Sales[Sales])

 

Best Regards,
Adamk Kong

 

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

  • yasbos's avatar
    yasbos
    Resolver II
    2 years ago

    Thanks so much Anonymous . I forgot to mention that I did know this solution, but I was hoping for a more localized decision because the calculate function that I have is much larger than just one line. I was hoping to modularize the code. Thanks so much for your help, nonetheless.

  • yasbos's avatar
    yasbos
    Resolver II
    2 years ago

    Also, thank you again so much for taking the time and putting the efforts into this, Anonymous . That's awesome!