Forum Discussion
How do an average for a column type date
The returned values are false because i have a result = 2089/09/16
Can you add 'NumDays' to this example, i'm curious what your 5 digit dates look like...???
- AlexGallet019 years agoHelper IV
NumDays = DATEDIFF(0;IF(TECHNIKE[THCJDATE]>0;TECHNIKE[THCJDATE];0);DAY)
- AlexGallet019 years agoHelper IV
DATE MOY = FORMAT(DATE(2000;1;INT(AVERAGE(TECHNIKE[NumDays])-36523));"YYYY-MM-DD")
- AlexGallet019 years agoHelper IV
- fhill9 years agoResident Rockstar
Here is the issue... These values are wrong... I'm not sure how your 'Date MOY' is being caculated (or where it's from) but 2015-12-08 and 2015-12-06 should only be 2 NUMDays apart, not 230,000+
- AlexGallet019 years agoHelper IV
I try to modify the colum but i have an error and i don't understand why i have this error
- v-yulgu-msft9 years agoMicrosoft Employee
Hi AlexGallet01,
What does the error message point to? Could you please translate it to English?
Regards,
Yuliana gu - AlexGallet019 years agoHelper IV
Hi,
My start date are above my end date
Or it's impossible because my start date are 0
- v-yulgu-msft9 years agoMicrosoft Employee
Hi AlexGallet01,
To use DateDiff function, we should meet the syntax that "the start date cannot be greater than end date". Otherwise, it will prompt error.
Regards,
Yuliana Gu