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Anonymous
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4 years ago
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Labour Availability

Hi,

 

I am using the Custom Calendar visual by Akvelon to show a labour profile of available labour on any given day in the month.

 

Dataset 1
My first dataset includes a list of every employee name and employee number (these employees should be on site every day)

 

Dataset 2
My second dataset includes employee number and start and end dates for sickness/holiday.

 

Please can someone explain how to create a measure to show the following;

 

Calendar visual showing all employees available on any given day (Logic = count of all employees in dataset 1 - employees who have dates in dataset 2)

 

I have attached an example report visual.

 

Many thanks,

  • Hi, Anonymous ;

    You could use custom calendar

    unfortunately it don't have tooltip. like below:

    1.create a calendar table.

    Table = CALENDAR(DATE(2021,1,1),DATE(2021,12,31))

    2.create a measure.

    Measure = 
    DISTINCTCOUNT('Table1'[employee name])-
    CALCULATE(DISTINCTCOUNT(Table2[employee number]),FILTER(ALL(Table2),[start date]<=MAX('Table'[Date])&&[end date]>=MAX('Table'[Date])))&"  available"

    The final output is shown below:

     

    Best Regards,
    Community Support Team_ Yalan Wu
    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

1 Reply

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

    Hi, Anonymous ;

    You could use custom calendar

    unfortunately it don't have tooltip. like below:

    1.create a calendar table.

    Table = CALENDAR(DATE(2021,1,1),DATE(2021,12,31))

    2.create a measure.

    Measure = 
    DISTINCTCOUNT('Table1'[employee name])-
    CALCULATE(DISTINCTCOUNT(Table2[employee number]),FILTER(ALL(Table2),[start date]<=MAX('Table'[Date])&&[end date]>=MAX('Table'[Date])))&"  available"

    The final output is shown below:

     

    Best Regards,
    Community Support Team_ Yalan Wu
    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.