DateAndTime.DateDiff Method

Source:

DateAndTime.vb

Source:

DateAndTime.vb

Source:

DateAndTime.vb

Source:

DateAndTime.vb

public static long DateDiff(Microsoft.VisualBasic.DateInterval Interval, DateTime Date1, DateTime Date2, Microsoft.VisualBasic.FirstDayOfWeek DayOfWeek = Microsoft.VisualBasic.FirstDayOfWeek.Sunday, Microsoft.VisualBasic.FirstWeekOfYear WeekOfYear = Microsoft.VisualBasic.FirstWeekOfYear.Jan1);

static member DateDiff : Microsoft.VisualBasic.DateInterval * DateTime * DateTime * Microsoft.VisualBasic.FirstDayOfWeek * Microsoft.VisualBasic.FirstWeekOfYear -> int64

Public Function DateDiff (Interval As DateInterval, Date1 As DateTime, Date2 As DateTime, Optional DayOfWeek As FirstDayOfWeek = Microsoft.VisualBasic.FirstDayOfWeek.Sunday, Optional WeekOfYear As FirstWeekOfYear = Microsoft.VisualBasic.FirstWeekOfYear.Jan1) As Long

Parameters

Returns

A long value specifying the number of time intervals between two Date values.

Exceptions

Examples

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

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.

  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

See also

• DateAdd
• DatePart
• Day(DateTime)
• Format(Object, String)
• Now
• Weekday(DateTime, FirstDayOfWeek)
• Year(DateTime)
• DateTime
• TimeSpan
• Data Type Summary (Visual Basic)

Source:

DateAndTime.vb

Source:

DateAndTime.vb

Source:

DateAndTime.vb

Source:

DateAndTime.vb

public static long DateDiff(string Interval, object? Date1, object? Date2, Microsoft.VisualBasic.FirstDayOfWeek DayOfWeek = Microsoft.VisualBasic.FirstDayOfWeek.Sunday, Microsoft.VisualBasic.FirstWeekOfYear WeekOfYear = Microsoft.VisualBasic.FirstWeekOfYear.Jan1);

public static long DateDiff(string Interval, object Date1, object Date2, Microsoft.VisualBasic.FirstDayOfWeek DayOfWeek = Microsoft.VisualBasic.FirstDayOfWeek.Sunday, Microsoft.VisualBasic.FirstWeekOfYear WeekOfYear = Microsoft.VisualBasic.FirstWeekOfYear.Jan1);

static member DateDiff : string * obj * obj * Microsoft.VisualBasic.FirstDayOfWeek * Microsoft.VisualBasic.FirstWeekOfYear -> int64

Public Function DateDiff (Interval As String, Date1 As Object, Date2 As Object, Optional DayOfWeek As FirstDayOfWeek = Microsoft.VisualBasic.FirstDayOfWeek.Sunday, Optional WeekOfYear As FirstWeekOfYear = Microsoft.VisualBasic.FirstWeekOfYear.Jan1) As Long

Parameters

Returns

A long value specifying the number of time intervals between two Date values.

Exceptions

Examples

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

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.

  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

See also

• DateAdd
• DatePart
• Day(DateTime)
• Format(Object, String)
• Now
• Weekday(DateTime, FirstDayOfWeek)
• Year(DateTime)
• DateTime
• TimeSpan
• Data Type Summary (Visual Basic)