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Anonymous
Not applicable

Calculate measures on the basis of 2 related tables but ignoring this relationship?

Hi everyone!

 

I have 2 tables: one table is a fact table containing dates field and the number of events, another table is a dimension table containing names and dates of creating these names. These tables are related by the field id_n, but I need calculate measure that will calculate the number of names (from table 2) per each date (from table 1) where created_date (from table 2) is greater than or equal to date (from table 1). 

 

The formula should looks like measure_table_1 = COUNT( IF(table_2[created_date] > = table_1[date], table_2[id_n]) )

 

Screenshot 2021-05-07 110057.png

 

Thanks!

1 ACCEPTED SOLUTION

Hi @Anonymous ,

You can create a measure as below, please find the details in the attachment.

Count of names = 
CALCULATE (
    DISTINCTCOUNT ( 'Table 2'[name_id] ),
    FILTER (
        ALL('Table 2'),
        'Table 2'[create_date] >= SELECTEDVALUE ( 'Table 1'[date] )
    )
)

yingyinr_1-1620723092415.png

Best Regards

Community Support Team _ Rena
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

4 REPLIES 4
amitchandak
Super User
Super User

@Anonymous , Try like

measure_table_1 = calculate(COUNTX(table_2, IF(table_2[created_date] > = max(table_1[date]), table_2[id_n],blank()) ),values(table_2[id_n]), crossfilter(table_2[id_n],table_1[id_n], none))

Anonymous
Not applicable

Hi. Thanks for the answer. But is it possible to do calculation without creating a new table?

 

Hi @Anonymous ,

You can create a measure as below, please find the details in the attachment.

Count of names = 
CALCULATE (
    DISTINCTCOUNT ( 'Table 2'[name_id] ),
    FILTER (
        ALL('Table 2'),
        'Table 2'[create_date] >= SELECTEDVALUE ( 'Table 1'[date] )
    )
)

yingyinr_1-1620723092415.png

Best Regards

Community Support Team _ Rena
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable

Thank you very much!

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