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Column caluclation with multiple cross table filteres

Hi Community

I have a problem to calculate the following. In one data source I have all my revenues splitted by quarter. In another table I have a list with margin per quarter. Now I would like to calculate in my revenue table the effictive margin related to the margin for the quarter.

For example:

 Table 1 Product Quarter Revenue Margin(Calc) A Q2 500 ? B Q3 300 ? C Q4 600 ?

 Table 2 Product Margin Quartal A 0.2 Q2 B 0.3 Q2 C 0.4 Q2 A 0.2 Q3 B 0.3 Q3 C 0.4 Q3 A 0.2 Q4 B 0.3 Q4 C 0.4 Q4

Thank you very much for your help!

1 ACCEPTED SOLUTION

Accepted Solutions
Community Support

Re: Column caluclation with multiple cross table filteres

Hi @Anubis0815 ,

You may create column or measure in Table1 like DAX below.

```Column: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = Table1[Product]&& Table2[Quartal] = Table1[Quarter]))

Measure: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = MAX(Table1[Product])&& Table2[Quartal] = MAX(Table1[Quarter])))```

Or you can also try to merge the two tables on the columns Product and Quarter in Query Editor.

Best Regards,

Amy

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

Community Support

Re: Column caluclation with multiple cross table filteres

Hi @Anubis0815 ,

You may create column or measure in Table1 like DAX below.

```Column: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = Table1[Product]&& Table2[Quartal] = Table1[Quarter]))

Measure: Margin(Calc)= CALCULATE (FIRSTNONBLANK ( Table2[Margin], 1 ), FILTER ( ALL ( Table2 ), Table2[Product] = MAX(Table1[Product])&& Table2[Quartal] = MAX(Table1[Quarter])))```

Or you can also try to merge the two tables on the columns Product and Quarter in Query Editor.

Best Regards,

Amy

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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