DateDiff Function (Visual Basic)
Returns a Long value specifying the number of time intervals between two Date values.
Public Overloads Function DateDiff( _ ByVal Interval As [ DateInterval | String ], _ ByVal Date1 As DateTime, _ ByVal Date2 As DateTime, _ Optional ByVal DayOfWeek As FirstDayOfWeek = FirstDayOfWeek.Sunday, _ Optional ByVal WeekOfYear As FirstWeekOfYear = FirstWeekOfYear.Jan1 _ ) As Long
The Interval argument can have one of the following settings.
Unit of time difference
The DayOfWeek argument can have one of the following settings.
First day of week specified in system settings
Monday (complies with ISO standard 8601, section 3.17)
The WeekOfYear argument can have one of the following settings.
First week of year specified in system settings
Week in which January 1 occurs (default)
Week that has at least four days in the new year (complies with ISO standard 8601, section 3.17)
First full week in the new year
Date1, Date2, or DayofWeek is out of range.
Date1 or Date2 is of an invalid type.
See the "Error number" column if you are upgrading Visual Basic 6.0 applications that use unstructured error handling. (You can compare the error number against the Number Property (Err Object).) However, when possible, you should consider replacing such error control with Structured Exception Handling Overview for Visual Basic.
You can use the DateDiff function to determine how many specified time intervals exist between two date/time values. 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.
Treatment of Parameters. DateDiff subtracts the value of Date1 from the value of Date2 to produce the difference. Neither value is changed in the calling program.
Return Values. Because Date1 and Date2 are of the Date data type, they hold date and time values accurate to 100-nanosecond ticks on the system timer. However, DateDiff always returns the number of time intervals as a Long value.
If Date1 represents a later date and time than Date2, DateDiff returns a negative number.
Day Intervals. If Interval is set to DateInterval.DayOfYear, it is treated the same as DateInterval.Day, because DayOfYear is not a meaningful unit for a time interval.
Week Intervals. If Interval is set to DateInterval.WeekOfYear, the return value represents the number of weeks between the first day of the week containing Date1 and the first day of the week containing Date2. The following example shows how this produces different results from DateInterval.Weekday.
' The following statements set datTim1 to a Thursday ' and datTim2 to the following Tuesday. Dim datTim1 As Date = #1/4/2001# Dim datTim2 As Date = #1/9/2001# ' Assume Sunday is specified as first day of the week. Dim wD As Long = DateDiff(DateInterval.Weekday, datTim1, datTim2) Dim wY As Long = DateDiff(DateInterval.WeekOfYear, datTim1, datTim2)
In the preceding example, DateDiff returns 0 to wD because the difference between the two dates is less than seven days, but it returns 1 to wY because there is a seven-day difference between the first days of the respective calendar weeks.
When the time part of Date1 is greater than that of Date2 and Interval is set to DateInterval.WeekOfYear, the DateDiff function returns a value that is one less than the correct value.
Larger Intervals. If Interval is set to DateInterval.Year, the return value is calculated purely from the year parts of Date1 and Date2. Similarly, the return value for DateInterval.Month is calculated purely from the year and month parts of the arguments, and for DateInterval.Quarter from the quarters containing the two dates.
For example, when comparing December 31 to January 1 of the following year, DateDiff returns 1 for DateInterval.Year, DateInterval.Quarter, or DateInterval.Month, even though at most only one day has elapsed.
Other Intervals. Since every Date value is supported by a DateTime structure, its methods give you additional options in finding time intervals. For example, you can use the Subtract method in either of its overloaded forms: DateTime.Subtract subtracts a TimeSpan from a Date variable to return another Date value, and DateTime.Subtract subtracts a Date value to return a TimeSpan. You can time a process to find out how many milliseconds it takes, as the following example shows.
This example uses the DateDiff function to display the number of days between a given date and today.
Assembly: Visual Basic Runtime Library (in Microsoft.VisualBasic.dll)