Forum Discussion
Getting 12 Months Running Total
- 6 years ago
Hi Anonymous
Create a measure
Measure = CALCULATE ( SUM ( 'Fact'[Value] ), FILTER ( ALLSELECTED ( 'Fact' ), 'Fact'[related date] <= MAX ( 'Fact'[related date] ) && DATEDIFF ( 'Fact'[related date], MAX ( 'Fact'[related date] ), MONTH ) <= 12 ) )Best Regards
Maggie
Community Support Team _ Maggie Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
There is a Running Total quick measure. Also, see if my Time Intelligence the Hard Way provides a different way of accomplishing what you are going for.
https://community.powerbi.com/t5/Quick-Measures-Gallery/Time-Intelligence-quot-The-Hard-Way-quot-TITHW/m-p/434008
- Anonymous6 years agoNot applicable
Thanks for your reply Greg, the quick measure did not worked and I created another one my self using different option both gave me same incorrect result.
Quick Measure =CALCULATE([Sum of value],FILTER(ALLSELECTED('Period'[Period]),ISONORAFTER('Period'[Period], MAX('Period'[Period]), DESC)))Using my DaxRunning Total =CALCULATE([Sum of value],DATESBETWEEN(Period[FirstDayOfMonth],MINX(ALL(Period),Period[FirstDayOfMonth]),MAX(Period[FirstDayO- Anonymous6 years agoNot applicable
following measure gives me Running total for continious period of time, Can some one help me to modify it for getting running total upto last 12 months only?
Running Total =CALCULATE([Sum of value],FILTER(ALL(Period),Period[FirstDayOfMonth] <= MAX (Period[FirstDayOfMonth])),VALUES(Period[FirstDayOfMonth].[Year]),VALUES(Period[FirstDayOfMonth].[Month]))- v-juanli-msft6 years agoCommunity Support
Hi Anonymous
Create a measure
Measure = CALCULATE ( SUM ( 'Fact'[Value] ), FILTER ( ALLSELECTED ( 'Fact' ), 'Fact'[related date] <= MAX ( 'Fact'[related date] ) && DATEDIFF ( 'Fact'[related date], MAX ( 'Fact'[related date] ), MONTH ) <= 12 ) )Best Regards
Maggie
Community Support Team _ Maggie Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.