Forum Discussion
HELP - Calculated Column Sales Between Dates For Each Row
- 3 years ago
Hi, VazhaPBI
Try this function. There is no relationship between the two tables.
Sales Amount = CALCULATE ( SUM ( 'Table B'[Sales Amount] ), FILTER ( 'Table B', [Date] >= EARLIER ( 'Table A'[Start Date] ) && [Date] <= EARLIER ( 'Table A'[End Date] ) &&[ClientID]=EARLIER('Table A'[ClientID]) ) )Best Regards,
Community Support Team _Charlotte
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Hi, VazhaPBI
You can try the following methods.
Column:
Sales Amount =
CALCULATE (
SUM ( 'Table B'[Sales Amount] ),
FILTER (
'Table B',
[Date] >= EARLIER ( 'Table A'[Start Date] )
&& [Date] <= EARLIER ( 'Table A'[End Date] )
)
)
Is this the result you expect?
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.
Thank You, for reply but I still have same problem, uploading some more screenshots. i cant understand what im doing wrong.
- v-zhangti3 years agoCommunity Support
Hi, VazhaPBI
Note that no relationship is required between the two calculated tables.
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.
- VazhaPBI3 years agoFrequent Visitor
But how expression understands without relationship, which client ID to sum up, if I have several other client iD,s in data, if not relationship than i have to use CLientID = CLientID and I still get same result. I just need it to make sum for every row independently, but what it does is just sum on one row with earlier start date.
- v-zhangti3 years agoCommunity Support
Hi, VazhaPBI
Try this function. There is no relationship between the two tables.
Sales Amount = CALCULATE ( SUM ( 'Table B'[Sales Amount] ), FILTER ( 'Table B', [Date] >= EARLIER ( 'Table A'[Start Date] ) && [Date] <= EARLIER ( 'Table A'[End Date] ) &&[ClientID]=EARLIER('Table A'[ClientID]) ) )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.