в 16:33, 26 мая 2010

2010-05-26T16:33:00Z

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2010-05-26T12:48:50Z

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==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>

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в 16:33, 26 мая 2010

Admin: 1 версия