Forum Discussion
Column Values Comparison
- 3 years ago
Hi, SamuelROS
Please try calculated column like:
Result = VAR _count = CALCULATE ( DISTINCTCOUNT('Table'[Interim]), FILTER ( ALL ( 'Table' ), 'Table'[Leave Date] = EARLIER ( 'Table'[Leave Date] ) && 'Table'[Leave Type] = "Leave" && 'Table'[Interim] <> EARLIER ( 'Table'[E Name] ) ) ) RETURN IF ( _count > 0, "conflicts" )If it doesn't work, please share your expected output in excel.
Best Regards,
Community Support Team _ Eason
SamuelROS , 17th also has the same issue, why not that ?
23rd is only one record
Hi amitchandak ,
yes so basically I'm after all records that has the same date && values in Ename can be found in the record of any of the same date's Interim column. What i'm trying to achieve is to determine if a user's [E Name] leave date conflicts with his Interim [Interim].
- v-easonf-msft3 years agoCommunity Support
Hi, SamuelROS
Please try calculated column like:
Result = VAR _count = CALCULATE ( DISTINCTCOUNT('Table'[Interim]), FILTER ( ALL ( 'Table' ), 'Table'[Leave Date] = EARLIER ( 'Table'[Leave Date] ) && 'Table'[Leave Type] = "Leave" && 'Table'[Interim] <> EARLIER ( 'Table'[E Name] ) ) ) RETURN IF ( _count > 0, "conflicts" )If it doesn't work, please share your expected output in excel.
Best Regards,
Community Support Team _ Eason- SamuelROS3 years agoFrequent Visitor
Hi CST,
It didn't work the way I wanted it to since it was returning the same values. But I did revised it a bit and it work like a charm :). Thank you for taking the time to look into my problem. Your code was also very easy to understand.
Result = VAR _count = CALCULATE ( DISTINCTCOUNT('Table'[Leave Date]), FILTER ( ALL ( 'Table' ), 'Table'[Leave Date] = EARLIER ( 'Table'[Leave Date] ) && 'Table'[Interim] = EARLIER ( 'Table'[E Name] ) ) ) RETURN IF ( _count > 0, "Conflict","No Conflict" )