Forum Discussion
Days Difference between work order previous open date and last work order closed date
Hi All,
I have a query as i have a table where i wish to create a difference between the same serial number's first Work order closed and second Work order opened date as Days Diff so that i can find out when the next work order was created after the previous one was closed . Please support . Below is the list of details. i have lots of serial numbers and Work order none in a sequence but randoms.
Hi Anonymous ,
Please add an index column in Power Query:
Then please try following DAX:
Days difference = var cur_index = 'Table'[Index] var pre_index = cur_index+1 var diffe = CALCULATE( MAX('Table'[CreateDate]), FILTER('Table','Table'[Index] = pre_index && 'Table'[Serial Number] = EARLIER('Table'[Serial Number]))) return DATEDIFF([ClosedOn],diffe,DAY)If I misunderstand your demands, please feel free to contact us in time.
Best regards,
Yadong Fang
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 , a new column
Datediff( [Close Date], Minx(filter(Table, [Serial Number]= earlier([Searial Number]) && [Open Date] > earlier([Close date]) ) , [Open Date]) , Day)
or
Datediff( [Close Date], Minx(filter(Table, [Serial Number]= earlier([Searial Number]) && [Open Date] > earlier([Open date]) ) , [Open Date]) , Day)- AnonymousNot applicable
I think its not working could you confirm :
- v-yadongf-msftCommunity Support
Hi Anonymous ,
Please add an index column in Power Query:
Then please try following DAX:
Days difference = var cur_index = 'Table'[Index] var pre_index = cur_index+1 var diffe = CALCULATE( MAX('Table'[CreateDate]), FILTER('Table','Table'[Index] = pre_index && 'Table'[Serial Number] = EARLIER('Table'[Serial Number]))) return DATEDIFF([ClosedOn],diffe,DAY)If I misunderstand your demands, please feel free to contact us in time.
Best regards,
Yadong Fang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.