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Anonymous's avatar
Anonymous
Not applicable
7 years ago
Solved

Qualifying Date for Date-Range

Hello all   I have a table of objects. These objects are available within a data range. dateActiveOn = active since dateDeadOn = inactive from   I now need to set a qualifying date to show o...
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  • v-juanli-msft's avatar
    v-juanli-msft
    7 years ago

    Hi Anonymous

    Explaination for the formula as below

    qualifying1 =
IF (
    [selected date] < MAX ( Sheet5[datedeadon] ) //"max" -> get the current row of a column,
        && [selected date] >= MAX ( Sheet5[dateactiveon] ),
    1,
    0
)

     

     

    I make a new test as below

     

    in my test, [DeadActiveOn] is date/time type, DateDeadOn and Date (Date table) are in date type.

     

    Could you share a screenshot to let me know your data and what's error of the qualifying measure?

     

     

    Best Regards

    Maggie

v-juanli-msft's avatar
v-juanli-msft
Community Support
7 years ago

Hi Anonymous

Create a date table which is not connected to your data table.

Date table =
ADDCOLUMNS (
    CALENDARAUTO (),
    "year", YEAR ( [Date] ),
    "month", MONTH ( [Date] ),
    "day", DAY ( [Date] ),
    "weeknum", WEEKNUM ( [Date], 2 ),
    "weekday", WEEKDAY ( [Date], 2 )
)

 

scenario 1: select one date from the "date table"

create measures in your main data table

selected date = SELECTEDVALUE('Date table'[Date])

qualifying1 =
IF (
    [selected date] < MAX ( Sheet5[datedeadon] )
        && [selected date] >= MAX ( Sheet5[dateactiveon] ),
    1,
    0
)

add "year","month","day" from the "date table" in the slicers,

add measure "qualifying" in Visual level filter.

 

 

scenario 2: select a date period from the "date table"

create measures in your main data table

 

min = MIN('Date table'[Date])

max = MAX('Date table'[Date])

qualifying2 =
IF (
    [min] >= MAX ( Sheet5[dateactiveon] )
        && [max] < MAX ( Sheet5[datedeadon] ),
    1,
    0
)

 

 

Best Regards

Maggie

 

Community Support Team _ Maggie Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

  • Anonymous's avatar
    Anonymous
    Not applicable
    7 years ago

    Hello v-juanli-msft

     

    Thanks for this answer.

    Im looking to get scenario 1 to work.

    Its about the same direction i tried yesterday night, but with a few changes...

     

    For now, i dont get it to work.

    Following Questions:

    • If DateDeadOn, DeadActiveOn and Date (Date table) may have different formats i.e Date with time or without. Does this matter?
    • In the formula, its possible to say DateDeadOn.Date  -- Should i use Date?
    • The 2 above doesnt really matter i think, the qualifying formula wont work this way. Why use Max?
      If the object hasnt a specified deadOn date its 12.31.9999, max is getting this date isnt it?

    The qualifying measure is always false, cant get it to true

    • v-juanli-msft's avatar
      v-juanli-msft
      Community Support
      7 years ago

      Hi Anonymous

      Explaination for the formula as below

      qualifying1 =
IF (
    [selected date] < MAX ( Sheet5[datedeadon] ) //"max" -> get the current row of a column,
        && [selected date] >= MAX ( Sheet5[dateactiveon] ),
    1,
    0
)

       

       

      I make a new test as below

       

      in my test, [DeadActiveOn] is date/time type, DateDeadOn and Date (Date table) are in date type.

       

      Could you share a screenshot to let me know your data and what's error of the qualifying measure?

       

       

      Best Regards

      Maggie

  • Anonymous's avatar
    Anonymous
    Not applicable
    7 years ago

    Hello v-juanli-msft

     

    With my inital matrix-table it showes 15k objects less than it should be. Because of that i thought it just dont work.

     

    I made a new table with the ID, activeOn, deadOn and qualifying. When i count the qualifyings (true) i get the correct number.

     

    So your solution works, i just need to implement it into the matrix correctly.