Forum Discussion
calculate difference in date/time based on multiple up and down time
- 2 years ago
you can try this
Column =VAR _last=maxx(FILTER('Table','Table'[DateTime]<EARLIER('Table'[DateTime])),'Table'[DateTime])VAR _lasta=maxx(FILTER('Table','Table'[DateTime]=_last),'Table'[Availability])VAR _last2=maxx(FILTER('Table','Table'[DateTime]<EARLIER('Table'[DateTime])&&'Table'[Availability]="U"),'Table'[DateTime])VAR _lasttime=minx(FILTER('Table','Table'[DateTime]>_last2),'Table'[DateTime])return if ('Table'[Availability]="U" && _lasta="U", blank(), if('Table'[Availability]="U" && ISBLANK(_lasttime),'Table'[DateTime]-max('Table'[DateTime]),if('Table'[Availability]="U",'Table'[DateTime]-_lasttime))) - Anonymous2 years ago
Hi EZiamslow ,
Based on the description, after getting data, selecting transform before Loading.
Selecting custom column and adding a year column.
Date.Year([Date]))Selecting custom column, add a month column.
Date.Month([Date])Then, selecting the desired year, reducing the number of data.
Closing and applying.
Then, using the DAX formula provided above.
Best Regards,
Wisdom Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi EZiamslow ,
Based on the description, after getting data, selecting transform before Loading.
Selecting custom column and adding a year column.
Date.Year([Date]))
Selecting custom column, add a month column.
Date.Month([Date])
Then, selecting the desired year, reducing the number of data.
Closing and applying.
Then, using the DAX formula provided above.
Best Regards,
Wisdom Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
which DAX formula are you referring to?
- Anonymous2 years agoNot applicable
Hi EZiamslow ,
Reducing the number of rows through the power query editor before loading the data. Then, using the ryan_mayu provide Dax formula. It should help.
Best Regards,
Wisdom Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.