Forum Discussion
Date/Time Difference between ID
- 3 years ago
Hi qwertzuiop ,
Please try:
TimeDiff_from_FirstTime_to_LastTime = var _a = MAXX(FILTER(ALL('Table'),[ID]=SELECTEDVALUE('Table'[ID])&&[CheckOccurence]="FirstTime"),[Date/time]) var _b = MAXX(FILTER(ALL('Table'),[ID]=SELECTEDVALUE('Table'[ID])&&[CheckOccurence]="LastTime"),[Date/time]) return DATEDIFF(_a,_b,MINUTE)/(60*24)Final output:
Best Regards,
Jianbo Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
qwertzuiop Please share your ENDOUTPUT in table for more understanding.
Hi Mahesh0016
Hopefully with this visualisation I can better represent my desired output.
It is not necessarily important that the output is in the corresponding line. What is important is that somewhere there is information that, for example, ID MDY5MP takes approx. 3,999 days from first to lasttime (maybe best approach in a measure, no?)
Let me know if you need further information.
Thank you
Cheers qwertzuiop
- v-jianboli-msft3 years agoCommunity Support
Hi qwertzuiop ,
Please try:
TimeDiff_from_FirstTime_to_LastTime = var _a = MAXX(FILTER(ALL('Table'),[ID]=SELECTEDVALUE('Table'[ID])&&[CheckOccurence]="FirstTime"),[Date/time]) var _b = MAXX(FILTER(ALL('Table'),[ID]=SELECTEDVALUE('Table'[ID])&&[CheckOccurence]="LastTime"),[Date/time]) return DATEDIFF(_a,_b,MINUTE)/(60*24)Final output:
Best Regards,
Jianbo Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.