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Regnbagar1's avatar
Regnbagar1
Icon for Helper I rankHelper I
4 years ago
Solved

Create a calculated column for a specifik date range

Hi, I need help with creating a calculated column in my date table, "Date_tabel". I would like to have 1 if it is between last 25:th and the 25:th before that otherwise 0.  For example: 1. Tod...
  • Anonymous's avatar
    Anonymous
    4 years ago

    Hi Regnbagar1 ,

    Please refer to my pbix file to see if it helps you. 

    Create a column.

    Column =
VAR _previous1 =
    DATE ( YEAR ( EDATE ( TODAY (), -1 ) ), MONTH ( EDATE ( TODAY (), -1 ) ), 25 )
VAR _previous2 =
    DATE ( YEAR ( EDATE ( TODAY (), -2 ) ), MONTH ( EDATE ( TODAY (), -2 ) ), 25 )
RETURN
    IF (
        'Table'[date] > _previous2
            && 'Table'[date] <= _previous1,
        1,
        IF ( TODAY () >= _previous1, 0, BLANK () )
    )

    If I have misunderstood your meaning, please provide your desired output and your pbix without privacy information.

     

    Best Regards

    Community Support Team _ Polly

     

     

Regnbagar1's avatar
Regnbagar1
Icon for Helper I rankHelper I
4 years ago

Hi,

 

Thank for the feedback. In my date table (see picture) I want to create a calculated dynamic table called "date range". The dates that are between the 25:th and 25:th the month before should have a 1 in "date range" column. In the example below the dates between 25:th of january and 25:th of february shoud have 1 in "date range" column and the rest of the dates 0 in "date range" column. 

Kind Regards

amitchandak's avatar
amitchandak
Icon for Super User rankSuper User
4 years ago

Regnbagar1 , try a new column like

if([Date] <= Today() ,if(Day([Date])<=25, datediff(eomonth([Date],-2) +26 ,datediff(eomonth([Date],-1) +26) ,today(),month), blank())

  • Regnbagar1's avatar
    Regnbagar1
    Icon for Helper I rankHelper I
    4 years ago

    Thank you so much. When I try the function I get the error message "Too few arguments were passed to the DATEDIFF function. The minimum argument count for the function is 3." 

     

    Do you know how I can handle that in the best way?