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Anonymous's avatar
Anonymous
Not applicable
6 years ago
Solved

Maximum Difference in Date Values

I have 5 different date value columns (each signifying the last scan date for 5 categories of devices). I want to create a column that would take the oldest date from these 5 columns and calculate the difference in days from TODAY. 

 

How should I do this?

 

 

  • Hi, Anonymous 

     

    Based on your description, I also suggest you use one date column. I created data t reproduce your scenario.

    Table:

     

    You may create two measures as follows.

     

    The oldest date = 
    CALCULATE(
        MIN('Table'[Date]),
        ALL('Table')
    )
    
    Difference = 
    DATEDIFF(
        [The oldest date],
        NOW(),
        DAY
    )

     

     

    Result:

     

    Best Regards

    Allan

     

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

3 Replies

  • Hi,

    Instead of maintaining 5 column, we should have all dates in 1 column.  Share some data and show the expected result.

  • I think the best way is, what suggested by Ashish_Mathur .

    You can try

     

    min date =
    var _tab = union(all(table[Date1]),all(table[Date2]),all(table[Date3]),all(table[Date4]),all(table[Date5]))
    Return
    minx(_tab,[date1])
    

     

    I have not tried it so Just check name suggest for date by pressing [  in last line

     

  • v-alq-msft's avatar
    v-alq-msft
    Community Support

    Hi, Anonymous 

     

    Based on your description, I also suggest you use one date column. I created data t reproduce your scenario.

    Table:

     

    You may create two measures as follows.

     

    The oldest date = 
    CALCULATE(
        MIN('Table'[Date]),
        ALL('Table')
    )
    
    Difference = 
    DATEDIFF(
        [The oldest date],
        NOW(),
        DAY
    )

     

     

    Result:

     

    Best Regards

    Allan

     

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