Forum Discussion
Calculate sum based on start and end date
- 2 years ago
Hi, mansi_luthra12
Add a year column to the date table and add a year filter to the formula.
Measure = CALCULATE ( SUM ( 'Table'[Value] ), FILTER ( ALL ( 'Table' ), MONTH ( [StartDateScope] ) <= SELECTEDVALUE ( 'Date'[Month] ) && MONTH ( [EndDateScope] ) >= SELECTEDVALUE ( 'Date'[Month] ) &&YEAR([StartDateScope])=SELECTEDVALUE('Date'[Year]) ) )If you do not get the results you expect, please provide more detailed data.
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.
Hi, mansi_luthra12
You can try the following methods.
Date table:
Measure =
CALCULATE ( SUM ( 'Table'[Value] ),
FILTER ( ALL ( 'Table' ),
MONTH ( [StartDateScope] ) <= SELECTEDVALUE ( 'Date'[Month] )
&& MONTH ( [EndDateScope] ) >= SELECTEDVALUE ( 'Date'[Month] )
)
)
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.
- mansi_luthra122 years agoNew Member
Hi,
Thanks for the reply the solution is perfectly fine if only one year is present. I have records for multiple years - from the calendar table even if I put a slicer on year it doesn't split the amount.
for Example : if Jan 2023 was 100 and 2024 Jan is 200 then irrespective of which year is selected the calculation shows 300.- v-zhangti2 years ago
Community Support
Hi, mansi_luthra12
Add a year column to the date table and add a year filter to the formula.
Measure = CALCULATE ( SUM ( 'Table'[Value] ), FILTER ( ALL ( 'Table' ), MONTH ( [StartDateScope] ) <= SELECTEDVALUE ( 'Date'[Month] ) && MONTH ( [EndDateScope] ) >= SELECTEDVALUE ( 'Date'[Month] ) &&YEAR([StartDateScope])=SELECTEDVALUE('Date'[Year]) ) )If you do not get the results you expect, please provide more detailed data.
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.