WebJan 18, 2024 · In the Filter tool, you would use the DateTimeDiff function. This returns an integer, so the expression might look something like... DateTimeDiff (OriginalDate, DerivedDate, "days") <= 30 Note: check the order of the above dates...they may need to be reversed to get what you are wanting. Reply 0 2 brandt3076 6 - Meteoroid 01-19-2024 … WebSep 17, 2024 · Max ( ( (DateTimeDiff ( [Service Time], [Arrival Time], "Seconds")/60)-1),0) I see in your data that the business logic throws away the first minute. It Report's in decimal minutes where you divide seconds by 60. If the results are less than 0, the answer is zero. Hopefully @tessaenns this helps. Cheers, Mark Alteryx ACE & Top Community Contributor
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WebMay 17, 2024 · The DateTimeDiff () calculation is literally counting the whole months between the occurrence of a date in the two month values. It is not counting the logical … WebApr 28, 2014 · The DateTimeDiff () function utilizes the Int32 data type on the back end and calculates hours and tries to take into account 'seconds' as well. Because of this, selecting 69 years in hours first gets calculated to 2,177,474,400 seconds, which is too large for an Int32 so it gets "wrapped around" to a negative number. camping ledrosee italien
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WebFeb 5, 2024 · The datetimediff () is going to round to the nearest whole number. What you can do, is calculate the difference in a smaller unit, and then divide to your unit of choice. In the example, I calculated the difference in seconds, then calculated the hours and days. Setting the datatype to fixed decimal with a scale of 2 will leave 2 decimal places. WebDec 11, 2024 · I am using Alteryx 10.5 and there are no DATEADD functions, only DATETIMEADD which returns a DATETIME type. Since my variables are both DATE … WebDec 6, 2016 · The dateTimeDiff function is another that uses different codes for parameters than other date functions or capabilities. I've created a suggestion to … camping le fief anduze