VBA/Excel/Access/Word/Date Functions/DateDiff — различия между версиями
Admin (обсуждение | вклад) м (1 версия) |
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Версия 19:33, 26 мая 2010
Содержание
- 1 DateDiff() function returns the difference between two dates. The unit for the difference is specified as a string ( "s" for second, "m" for month, "yyyy" for year, and so on).
- 2 DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]]) returns the interval between two specified dates
- 3 DateDiff("m", "0", Now)
- 4 DateDiff("m", Now, ")
- 5 DateDiff returns the interval of time between two dates:
- 6 DateDiff("yyyy", Now, ")
- 7 Difference between two dates in days and months:
- 8 returns the number of weeks between June 3, nd September:
- 9 Use DateDiff function
DateDiff() function returns the difference between two dates. The unit for the difference is specified as a string ( "s" for second, "m" for month, "yyyy" for year, and so on).
<source lang="vb">
Sub dateDiffDemo()
Debug.Print DateDiff("s", Now, "10/10/03")
End Sub
</source>
DateDiff(interval, date1, date2[, firstdayofweek[, firstweekofyear]]) returns the interval between two specified dates
<source lang="vb">
Constant Value Year Starts with Week vbUseSystem 0 Use the system setting. vbFirstJan1 1 The week in which January 1 falls (the default setting). vbFirstFourDays 2 The first week with a minimum of four days in the year. vbFirstFullWeek 3 The first full week (7 days) of the year.
</source>
DateDiff("m", "0", Now)
<source lang="vb">
Sub dateFunctions1()
Debug.Print "The months between 3/15/2000 and today is: " & DateDiff("m", "3/15/2000", Now)
End Sub
</source>
DateDiff("m", Now, ")
<source lang="vb">
Sub dateDiff2()
Debug.Print DateDiff("m", Now, "10/10/03")
End Sub
</source>
DateDiff returns the interval of time between two dates:
<source lang="vb">
Sub DateDiffExample()
Debug.Print DateDiff("d", Now, "12/31/2010") ""Days until 12/31/2010 Debug.Print DateDiff("m", Now, "12/31/2010") ""Months until 12/31/2010 Debug.Print DateDiff("yyyy", Now, "12/31/2010") ""Years until 12/31/2010 Debug.Print DateDiff("q", Now, "12/31/2010") ""Quarters until 12/31/2010
End Sub
</source>
DateDiff("yyyy", Now, ")
<source lang="vb">
Sub dateDiff3()
Debug.Print DateDiff("yyyy", Now, "10/10/03")
End Sub
</source>
Difference between two dates in days and months:
<source lang="vb">
Sub dateFunctions()
Dim strDateString As String strDateString = "The days between 3/15/2000 and today is: " & _ DateDiff("d", "3/15/2000", Now) & vbCrLf & _ "The months between 3/15/2000 and today is: " & _ DateDiff("m", "3/15/2000", Now) msgBox strDateString
End Sub
</source>
returns the number of weeks between June 3, nd September:
<source lang="vb">
Sub dateD()
MsgBox DateDiff("ww", "6/3/2006", "9/30/2006")
End Sub "Using the Dir Function to Check Whether a File Exists
Sub Does_File_Exist() Dim strTestFile As String, strNameToTest As String strNameToTest = "c:\" If strNameToTest = "" Then End strTestFile = Dir(strNameToTest) If Len(strTestFile) = 0 Then Debug.Print "The file " & strNameToTest & " does not exist." Else Debug.Print "The file " & strNameToTest & " exists." End If End Sub </source>
Use DateDiff function
<source lang="vb">
Sub dateDiffFunction()
Dim strDateString As String strDateString = "The days between 3/15/2000 and today is: " & _ DateDiff("d", "3/15/2000", Now) & vbCrLf & _ "The months between 3/15/2000 and today is: " & _ DateDiff("m", "3/15/2000", Now) MsgBox strDateString
End Sub
</source>