DateAndTime.DateDiff Method (String, Object, Object, FirstDayOfWeek, FirstWeekOfYear)

Returns a Long value specifying the number of time intervals between two Date values.

Namespace:  Microsoft.VisualBasic
Assembly:  Microsoft.VisualBasic (in Microsoft.VisualBasic.dll)

public static long DateDiff(
	string Interval,
	Object Date1,
	Object Date2,
	FirstDayOfWeek DayOfWeek,
	FirstWeekOfYear WeekOfYear
)

Parameters

Interval
Type: System.String

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
Type: System.Object

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

Date2
Type: System.Object

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

DayOfWeek
Type: Microsoft.VisualBasic.FirstDayOfWeek

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

WeekOfYear
Type: Microsoft.VisualBasic.FirstWeekOfYear

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

Return Value

Type: System.Int64
Returns a Long value specifying the number of time intervals between two Date values.

ExceptionCondition
ArgumentException

Invalid Interval.

ArgumentException

Date1, Date2, or DayofWeek is out of range.

InvalidCastException

Date1 or Date2 is of an invalid type.

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.

    For cultures such as Japanese that have multiple eras, the DateDiff method does not return a difference in years if the difference spans two or more eras. Instead, you can calculate the difference in values returned by the Year property, as shown in the following example: date2.Year - date1.Year.

  • 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.

    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 Double = runLength.TotalMilliseconds
    

The Interval argument can have one of the following settings.

Enumeration value

String value

Unit of time difference

DateInterval.Day

"d"

Day

DateInterval.DayOfYear

"y"

Day

DateInterval.Hour

"h"

Hour

DateInterval.Minute

"n"

Minute

DateInterval.Month

"m"

Month

DateInterval.Quarter

"q"

Quarter

DateInterval.Second

"s"

Second

DateInterval.Weekday

"w"

Week

DateInterval.WeekOfYear

"ww"

Calendar week

DateInterval.Year

"yyyy"

Year

The DayOfWeek argument can have one of the following settings.

Enumeration value

Value

Description

FirstDayOfWeek.System

0

First day of week specified in system settings

FirstDayOfWeek.Sunday

1

Sunday (default)

FirstDayOfWeek.Monday

2

Monday (complies with ISO standard 8601, section 3.17)

FirstDayOfWeek.Tuesday

3

Tuesday

FirstDayOfWeek.Wednesday

4

Wednesday

FirstDayOfWeek.Thursday

5

Thursday

FirstDayOfWeek.Friday

6

Friday

FirstDayOfWeek.Saturday

7

Saturday

The WeekOfYear argument can have one of the following settings.

Enumeration value

Value

Description

FirstWeekOfYear.System

0

First week of year specified in system settings

FirstWeekOfYear.Jan1

1

Week in which January 1 occurs (default)

FirstWeekOfYear.FirstFourDays

2

Week that has at least four days in the new year (complies with ISO standard 8601, section 3.17)

FirstWeekOfYear.FirstFullWeek

3

First full week in the new year

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

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. 
    Dim days As Long = DateDiff(DateInterval.Day, date1, date2)

    ' This statement has a string interval argument, and 
    ' is equivalent to the above statement. 
    'Dim days As Long = DateDiff("d", date1, date2)

    MessageBox.Show("Days from today: " & days.ToString)
Catch ex As Exception
    MessageBox.Show("Invalid Date: " & ex.Message)
End Try

.NET Framework

Supported in: 4.5, 4, 3.5, 3.0, 2.0, 1.1, 1.0

.NET Framework Client Profile

Supported in: 4, 3.5 SP1

Windows 8.1, Windows Server 2012 R2, Windows 8, Windows Server 2012, Windows 7, Windows Vista SP2, Windows Server 2008 (Server Core Role not supported), Windows Server 2008 R2 (Server Core Role supported with SP1 or later; Itanium not supported)

The .NET Framework does not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.

Was this page helpful?
(1500 characters remaining)
Thank you for your feedback
Show:
© 2014 Microsoft