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Anonymous
Not applicable

Discount count multiple tables

I have these two tables:

Assigment:

empid projid      

1011054
1011050
1031052
1071053
1021053
1041051

 

Projects:

projid  name   customer

1050SunPhilips
1051MoonASML
1052StarsASML
1053GalaxyDLL
1054MarsPhilips

 

As you can see, projid 1050 and 1054 and projid 1051 and 1052 have the same customer. I want to calculate the amount of projects an employee had for different customers, but if an employee has projid 1050 and 1054, it will count as 1 project. Does someone knows how to do this?

1 ACCEPTED SOLUTION
v-jayw-msft
Community Support
Community Support

Hi @mrstillchicken,

 

Here’s the sample data I used according to your description:

3.PNG4.PNG

Please check following steps as below and see if the result achieve your expectation:

1. Manage relationship between two tables:

1.PNG

2. Create measure:

    distinctcount = CALCULATE(DISTINCTCOUNT(Projects[customer]),FILTER(Assigment,Assigment[empid] = MAX(Assigment[empid])))

3. Result would be shown as below:

2.PNG

BTW, Pbix as attached. Hopefully works for you.

 

Best Regards,

Jay

 

Community Support Team _ Jay Wang

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

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

View solution in original post

1 REPLY 1
v-jayw-msft
Community Support
Community Support

Hi @mrstillchicken,

 

Here’s the sample data I used according to your description:

3.PNG4.PNG

Please check following steps as below and see if the result achieve your expectation:

1. Manage relationship between two tables:

1.PNG

2. Create measure:

    distinctcount = CALCULATE(DISTINCTCOUNT(Projects[customer]),FILTER(Assigment,Assigment[empid] = MAX(Assigment[empid])))

3. Result would be shown as below:

2.PNG

BTW, Pbix as attached. Hopefully works for you.

 

Best Regards,

Jay

 

Community Support Team _ Jay Wang

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

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

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