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

Parameters

• Interval
  Required. DateInterval enumeration value or String expression representing the time interval you want to use as the unit of difference between Date1 and Date2.

• Date1
  Required. Date. The first date/time value you want to use in the calculation.

• Date2
  Required. Date. The second date/time value you want to use in the calculation.

• DayOfWeek
  Optional. A value chosen from the FirstDayOfWeek enumeration that specifies the first day of the week. If not specified, FirstDayOfWeek.Sunday is used.

• WeekOfYear
  Optional. A value chosen from the FirstWeekOfYear enumeration that specifies the first week of the year. If not specified, FirstWeekOfYear.Jan1 is used.

Settings

The Interval argument can have one of the following settings.

The DayOfWeek argument can have one of the following settings.

The WeekOfYear argument can have one of the following settings.

Exceptions

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.

Remarks

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.

Behavior

• 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.

• 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: System.DateTime.Subtract(System.TimeSpan) subtracts a TimeSpan from a Date variable to return another Date value, and System.DateTime.Subtract(System.DateTime) 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.

  Dim startTime As Date = Now
  ' Run the process that is to be timed.
  Dim runLength As Global.System.TimeSpan = Now.Subtract(startTime)
  Dim millisecs As Integer = runLength.Milliseconds
  

Example

This example uses the DateDiff function to display the number of days between a given date and today.

Dim firstDate, msg As String
Dim secondDate As Date
firstDate = InputBox("Enter a date")
secondDate = CDate(firstDate)
msg = "Days from today: " & DateDiff(DateInterval.Day, Now, secondDate)
MsgBox(msg)

Requirements

Namespace: Microsoft.VisualBasic

Module: DateAndTime

Assembly: Visual Basic Runtime Library (in Microsoft.VisualBasic.dll)

See Also

Reference

DateAdd Function (Visual Basic)
DatePart Function (Visual Basic)
Day Function (Visual Basic)
Format Function
Now Property
Weekday Function (Visual Basic)
Year Function (Visual Basic)
Date Data Type (Visual Basic)
DateTime
TimeSpan