Forum Discussion
Latest Value Multiplication
Hello everybody,
I have a dataset similar to the below.
| Date | Job ID | Total Hours | % Complete |
| 01/01/2022 | 1 | 10 | 25 |
| 02/01/2022 | 1 | 10 | 50 |
| 03/01/2022 | 2 | 100 | 30 |
| 04/01/2022 | 2 | 100 | 100 |
I want to create a measure that can multiple column 'Total Hours' buy latest '% Complete' column per job ID.
For example
'Job ID' (1)
'Total Hours' (10) * '% Complete' (50) = 5
Any ideas on how to achieve this?
Many thanks,
- Anonymous4 years ago
Hi Anonymous ,
I have created a simple sample, please refer to it to see if it helps you.
Create a measure first.
Measure = VAR _maxdate = CALCULATE ( MAX ( 'Table'[Date] ), FILTER ( ALL ( 'Table' ), 'Table'[Job ID] = SELECTEDVALUE ( 'Table'[Job ID] ) ) ) RETURN IF ( MAX ( 'Table'[Date] ) = _maxdate, MAX ( 'Table'[Total Hours] ) * MAX ( 'Table'[% Complete ] ) / 100, BLANK () )If I have misunderstood your meaning, please provide more details with your desired output.
Best Regards
Community Support Team _ Polly
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly
3 Replies
- amitchandakSuper User
Anonymous , Create a column
= [hours] *[Percent complete] // Divide by 100 if need
then follow this blog for the measure
- AnonymousNot applicable
Thanks for your response. This is great but I only need the multiplication for the latest date that the ID has been mentioned.
- AnonymousNot applicable
Hi Anonymous ,
I have created a simple sample, please refer to it to see if it helps you.
Create a measure first.
Measure = VAR _maxdate = CALCULATE ( MAX ( 'Table'[Date] ), FILTER ( ALL ( 'Table' ), 'Table'[Job ID] = SELECTEDVALUE ( 'Table'[Job ID] ) ) ) RETURN IF ( MAX ( 'Table'[Date] ) = _maxdate, MAX ( 'Table'[Total Hours] ) * MAX ( 'Table'[% Complete ] ) / 100, BLANK () )If I have misunderstood your meaning, please provide more details with your desired output.
Best Regards
Community Support Team _ Polly
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly