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PawelTr
Regular Visitor

Group by several criteria and take oldest available value

Hello!

I'm struggling with DAX code.

To describe the problem, look at the picture below. I have many rows with the data, separated by invoice month, country, section, supplier type. In next four rows are some values, let's focus as example on Sum_selling_price

PawelTr_0-1683728624510.png


I want to create calculated column, which always take value for first available month(here it will be April) and assign in all 12 rows. 

My current code, where I failed looks as provided

PawelTr_1-1683728834777.png


I faced "circular dependency".

I would be really grateful for any help.



1 ACCEPTED SOLUTION
v-zhangti
Community Support
Community Support

Hi, @PawelTr 

 

You can try the following methods.

Selling price = 
Var _firstdate=CALCULATE(MIN('Table'[INVOICE_MONTH_END]),ALL('Table'))
Return
CALCULATE(SUM('Table'[Sum_selling_price]),FILTER('Table',[INVOICE_MONTH_END]=_firstdate))

vzhangti_0-1683874382994.png

Is this the result you expect? If not, please provide more information.

 

Best Regards,

Community Support Team _Charlotte

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

2 REPLIES 2
PawelTr
Regular Visitor

Hello v-zhangti,

Sorry for beeing silent for a long time. I wasn't able to log in recently.
Your advice helped me solved this problem. 

A veeery big thank you for this support! 😊

v-zhangti
Community Support
Community Support

Hi, @PawelTr 

 

You can try the following methods.

Selling price = 
Var _firstdate=CALCULATE(MIN('Table'[INVOICE_MONTH_END]),ALL('Table'))
Return
CALCULATE(SUM('Table'[Sum_selling_price]),FILTER('Table',[INVOICE_MONTH_END]=_firstdate))

vzhangti_0-1683874382994.png

Is this the result you expect? If not, please provide more information.

 

Best Regards,

Community Support Team _Charlotte

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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