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Anonymous's avatar
Anonymous
Not applicable
6 years ago
Solved

Count Of Days In Month.

Hi ,

Date                     hire date

05-Aug-2020       05-aug-19

 

I need how many months diffrence Date and hire date,please help on this issues.

 

 

Regards,

  • Anonymous , as new column

    datediff([hire date],today(),Month)

    or

    datediff([hire date],[date],Month)

     

    as a new measure // you need to take care of row context

    datediff(min([hire date]),today(),Month)

    or

    datediff(min([hire date]),Max([date]),Month)

4 Replies

  • Greg_Deckler's avatar
    Greg_Deckler
    Community Champion

    Anonymous - Maybe:

    Column =
      VAR __HireYear = YEAR([hire date])
      VAR __HireMonth = MONTH([hire date])
      VAR __DateYear = YEAR([Date])
      VAR __DateMonth = MONTH([Date])
    RETURN
      (__DateYear - __HireYear)*12 + (__HireMonth - __DateMonth)
  • az38's avatar
    az38
    Community Champion

    Hi Anonymous 

    try new column

    Column = DATEDIFF([hire date], [Date], MONTH)
  • Anonymous,

     

    Create a new column and use the DATEDIFF function.

    Here's a link to how DATEDIFF works: https://docs.microsoft.com/en-us/dax/datediff-function-dax

     

    For months difference between Hire_Date and Date, you will have to use below formula:

    Month_Diff Col = DATEDIFF([hire date], [Date], MONTH)

     

    For days difference between Hire_Date and Date, you will have to use below formula:

     

    Day_Diff Col = DATEDIFF([hire date], [Date], DAY)​

     

     

    Give a thumbs up if this post helped you in any way and mark this post as solution if it solved your query !!!

  • Anonymous , as new column

    datediff([hire date],today(),Month)

    or

    datediff([hire date],[date],Month)

     

    as a new measure // you need to take care of row context

    datediff(min([hire date]),today(),Month)

    or

    datediff(min([hire date]),Max([date]),Month)