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https://chennaiiq.com/developers/reference/visual_basic/functions/datediff.asp
DateDiff Function - Visual Basic 6.0 (VB 6.0) Returns a Variant (Long) specifying the number of time intervals between two specified dates. Syntax: DateDiff (interval, date1, date2 [, firstdayofweek [, firstweekofyear]]) Parameter. Description.
https://stackoverflow.com/questions/25153625/how-to-calculate-timedifference-in-vb6
For the purposes of getting your string time diff in hours:minutes:seconds, I would do: Dim lSeconds as Long, lMinutes as Long lSeconds = DateDiff("s", dFrom, dTo) lMinutes = Fix(lSeconds / 60) '// Gets the whole number (not rounded) lSeconds = ((lSeconds / 60) - lMinutes) * 60 '// get the remaining seconds sTimeDiff = "00:" & Format$(lMinutes, "00") & ":" & Format$(lSeconds, "00")
https://docs.microsoft.com/en-us/dotnet/api/microsoft.visualbasic.dateandtime.datediff
This example uses the DateDiff function to display the number of days between a given date and today. VB. Dim date2Entered As String = InputBox ("Enter a date") Try Dim date2 As Date = Date.Parse (date2Entered) Dim date1 As Date = Now ' Determine the number of days between the two dates.
https://www.w3schools.com/asp/func_datediff.asp
Two dates you want to use in the calculation. firstdayofweek. Optional. Specifies the day of the week. Can take the following values: 0 = vbUseSystemDayOfWeek - Use National Language Support (NLS) API setting. 1 = vbSunday - Sunday (default) 2 = vbMonday - Monday. 3 = vbTuesday - Tuesday.
https://www.fmsinc.com/MicrosoftAccess/modules/code/GeneralVBA_VB6/DateandTime/DateTime.htm
For instance, if you pass it 38959290 seconds, it will return "1 year, 2 months, 25 days, 2 hours, 1 minute, 30 seconds" based on 60 seconds per minute and 365 days per year. Note that the difference between 1/1/2005 and 3/26/2006 2:01:30 AM is 1 year, 2 months, 25 days, 2 hours, 1 minute, 30 seconds by calendar, and is 38800890 seconds using DateDiff("s", #1/1/2005#, #3/26/2006 2:01:30 …
https://social.msdn.microsoft.com/forums/sqlserver/en-US/8b602425-a3ed-4537-b929-d4ecdde3aa69/datediff-by-hour-and-minutes
DATEDIFF(mi. , [Start Time] , [Actual Start]))>0. I thought it would be easy just to do a datediff where it is greater than > 0. but i've noticed that the following dataset is still being pulled through... [planned start time] [actual start time] 1900-01-01 09:00:00.000 1900-01-01 09:00:19.000.
https://chennaiiq.com/developers/reference/visual_basic/functions/dateadd.asp
DateDiff Function DatePart Function Date Time Functions Visual Basic Functions Visual Basic Home : Description Syntax Return Datatype Example Migration. DateAdd Function - Visual Basic 6.0 (VB 6.0) Returns a Variant (Date) containing a date to which a specified time interval has been added. Syntax: DateAdd(interval, number, date) ...
https://support.microsoft.com/en-us/office/datediff-function-e6dd7ee6-3d01-4531-905c-e24fc238f85f
Remarks. You can use the DateDiff function to determine how many specified time intervals exist between two dates. For example, you might use DateDiff to calculate the number of days between two dates, or the number of weeks between today and the end of the year.. To calculate the number of days between date1 and date2, you can use either Day of year ("y") or Day ("d").
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