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Regular Visitor

## Sum of SubCategory

Hi, I have the following issue Please help,

I have a table with Month, Column A  and Column B

How do i get Column Total and sum that based on month.

1 ACCEPTED SOLUTION

Accepted Solutions
Established Member

## Re: Sum of SubCategory

@Mahadevaraobc

I am not sure if I understand your scenairo correclty.

You can use count. If you don't want to count duplicated the parts. You caan use distinctcount

Measure = COUNT(Sheet1[B])

Measure = DISTINCTCOUNT(Sheet1[B])

6 REPLIES 6
Established Member

## Re: Sum of SubCategory

@Mahadevaraobc

I think you can just use sum formula. Could you please explain what situation you are in?

`Measure = sum(Table3[Column1])`

Regular Visitor

## Re: Sum of SubCategory

Thank You,

I have some data in Column B, which is subcategory of column A, how can i count many are present and sum the value and show that for month.

Established Member

## Re: Sum of SubCategory

@Mahadevaraobc

Could you please provide more info and elaborate it? It's better to provide some sample data and the result you expected.Thanks.

Highlighted
Regular Visitor

## Re: Sum of SubCategory

Hi,
As I don't have sample data let me elaborate the issue
I have month column, column A ( serial no) and column B (part)
Eg . Column A has 10 serial numbers and assuming each serial no has multiple no parts in column B
Now how do I show in a month how many parts were sent for 1 serial number
Established Member

## Re: Sum of SubCategory

@Mahadevaraobc

I am not sure if I understand your scenairo correclty.

You can use count. If you don't want to count duplicated the parts. You caan use distinctcount

Measure = COUNT(Sheet1[B])

Measure = DISTINCTCOUNT(Sheet1[B])

Community Support Team

## Re: Sum of SubCategory

You could create some dummy data like what ryan_mayu did to illustrate your requirement.

How to Get Your Question Answered Quickly

Regards,

Yuliana Gu

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