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Hello,
I'm following Power BI tuto and good news, I'm already stuck at the TOTALYTD DAX function.
Here is an extract of the test table (Excel) I created (and I generated around 700 lines):
Sales | Date | MonthNumber | Year |
4 | 01/01/2016 | 1 | 2016 |
6 | 02/01/2016 | 1 | 2016 |
8 | 03/01/2016 | 1 | 2016 |
38 | 04/01/2016 | 1 | 2016 |
54 | 05/01/2016 | 1 | 2016 |
2 | 06/01/2016 | 1 | 2016 |
5 | 07/01/2016 | 1 | 2016 |
23 | 08/01/2016 | 1 | 2016 |
24 | 09/01/2016 | 1 | 2016 |
I'd like to have a cumulative view of the sales by month for each year so I created this variable:
YTD = TOTALYTD(SUM(Sheet1[Sales]);Sheet1[Date])
When I create a matrix with MonthNumbers as rows, Years as columns and YTD as values (right matrix), I get the same as when I create a matrix with MonthNumbers as rows, Years as columns and Sales as values (right matrix).
Does someone have an idea?
Thanks a lot!
Solved! Go to Solution.
Regards
Miguel Félix
Proud to be a Super User!
Check out my blog: Power BI em PortuguêsRegards
Miguel Félix
Proud to be a Super User!
Check out my blog: Power BI em PortuguêsHi @MFelix,
Thank you for your solution!
By creating a seperate calendar table and linking the 2 tables, my measures perfectly works.
Regards,
thimremy
Hi @thimremy
I dont get your point , u need to calculate cumulative sum of sales by month ?
if u neeed to calculate sum of sale group by month
measure= calculate(sum(sheet1[sales]); DATESMTD(Sheet1[Date])
Let me know your need exactly
Hi @Mariam1991,
Yes, I need to calculate a cumulative sum of sales througout the months (For example: for February 2017, I need to sum sales of January & February 2017).
As you said, to calculate sum of sales group by month, the following measure you offered perfectly works:
measure= calculate(sum(sheet1[sales]); DATESMTD(Sheet1[Date])
And to calculate a cumulative sum of sales througout the months, I tried @MFelix's solution to create a separate calendar table and I used my initial measure (YTD = TOTALYTD(SUM(Sales[Sales]);Date[Date])) and it perfectly worked.
Thank you for having answered so quickly!
Regards,
thimremy
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