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Anonymous
Not applicable

Calculate the number of weeks between the two date rows

Hi,

 

I need to calculate the number of weeks between the two date rows based on the category start date and finish date as shown below

 

 

sdc dates.PNG

1 ACCEPTED SOLUTION
Anonymous
Not applicable

Hi @Anonymous 

Try to use datediff function in your code.

Weekdiff = 
VAR _T =
    SUMMARIZE (
        'Table',
        'Table'[SDCCode],
        "Weekdiff",
            VAR _LoadStartDate =
                CALCULATE (
                    MAX ( 'Table'[Date] ),
                    FILTER ( 'Table', 'Table'[Category] = "03.WO Load Start Date" )
                )
            VAR _LoadFinishDate =
                CALCULATE (
                    MAX ( 'Table'[Date] ),
                    FILTER ( 'Table', 'Table'[Category] = "04.WO Load Finish Date" )
                )
            RETURN
                DATEDIFF ( _LoadStartDate, _LoadFinishDate, WEEK )
    )
RETURN
    SUMX ( _T, [Weekdiff] )

Result is as below.

1.png

Best Regards,
Rico Zhou

 

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

View solution in original post

1 REPLY 1
Anonymous
Not applicable

Hi @Anonymous 

Try to use datediff function in your code.

Weekdiff = 
VAR _T =
    SUMMARIZE (
        'Table',
        'Table'[SDCCode],
        "Weekdiff",
            VAR _LoadStartDate =
                CALCULATE (
                    MAX ( 'Table'[Date] ),
                    FILTER ( 'Table', 'Table'[Category] = "03.WO Load Start Date" )
                )
            VAR _LoadFinishDate =
                CALCULATE (
                    MAX ( 'Table'[Date] ),
                    FILTER ( 'Table', 'Table'[Category] = "04.WO Load Finish Date" )
                )
            RETURN
                DATEDIFF ( _LoadStartDate, _LoadFinishDate, WEEK )
    )
RETURN
    SUMX ( _T, [Weekdiff] )

Result is as below.

1.png

Best Regards,
Rico Zhou

 

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

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