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adam_macs
Helper I
Helper I

Count of Subcategory Rows related to another table

Hi, I have 2 tables that are connected via the Customer ID.

 

My aim is to have a distinct count of the total customers within a given subcategory of Table B if one or more customers from Table A are found within that subcategory.

 

This would tell me how much of the market we have penetrated. For example, we sell road bikes to Jon Yang which is 1 out of a possible 6 customers within the road bike category.

 

So essentially, if we have customers from Table A in Table B, what is the total number of customers within the subcategories they exist in.

 

Any help much appreciated.

 

adam_macs_1-1660560795748.png

1 ACCEPTED SOLUTION
Anonymous
Not applicable

Hi @adam_macs ,

 

I do the test and this is my data of two tables:

vmengzhumsft_1-1660811024283.png

I got this result through my testing:

vmengzhumsft_0-1660810728137.png

and through filtering, you can get the count of rows related to each subcategory.

for example, I choose Bike , it get the correct result:

vmengzhumsft_2-1660811068568.png

so if this result can meets your needs, you can try this measure:

 

Measure = COUNTROWS(FILTER('Table 1','Table 1'[customer id] in VALUES('Table 2'[customer id])))

 

 

Best regards,

Community Support Team Selina zhu

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

3 REPLIES 3
Anonymous
Not applicable

Hi @adam_macs ,

 

I do the test and this is my data of two tables:

vmengzhumsft_1-1660811024283.png

I got this result through my testing:

vmengzhumsft_0-1660810728137.png

and through filtering, you can get the count of rows related to each subcategory.

for example, I choose Bike , it get the correct result:

vmengzhumsft_2-1660811068568.png

so if this result can meets your needs, you can try this measure:

 

Measure = COUNTROWS(FILTER('Table 1','Table 1'[customer id] in VALUES('Table 2'[customer id])))

 

 

Best regards,

Community Support Team Selina zhu

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

 

jplouw1
New Member

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