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DaysInAnyMonth

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Submitted on: 2/1/2001 8:49:56 PM
By: Joe McDonnell 
Level: Beginner
User Rating: By 3 Users
Compatibility:ASP (Active Server Pages), VbScript (browser/client side)

Users have accessed this code 16303 times.
 
 
     Return the number of days in any month including leap years
 
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Terms of Agreement:   
By using this code, you agree to the following terms...   
1) You may use this code in your own programs (and may compile it into a program and distribute it in compiled format for languages that allow it) freely and with no charge.   
2) You MAY NOT redistribute this code (for example to a web site) without written permission from the original author. Failure to do so is a violation of copyright laws.   
3) You may link to this code from another website, but ONLY if it is not wrapped in a frame. 
4) You will abide by any additional copyright restrictions which the author may have placed in the code or code's description.

    '**************************************
    ' Name: DaysInAnyMonth
    ' Description:Return the number of days 
    '     in any month including leap years
    ' By: Joe McDonnell
    '
    ' Inputs:IntMonth .... number of month
    IntYear ..... 4 figure year or zero For current year
    '
    ' Returns:days in month
    '
    ' Assumes:A year is a leap year if:
    It is divisible by 4 but Not by 100, or
    It is divisible by 400.
    '
    'This code is copyrighted and has    ' limited warranties.Please see http://w
    '     ww.Planet-Source-Code.com/vb/scripts/Sho
    '     wCode.asp?txtCodeId=6477&lngWId=4    'for details.    '**************************************
    
    function DaysInAnyMonth(intMonth, IntYear)
    if IntYear = 0 Then
    IntYear = Year(Now())
    End if
    Select Case (intMonth)
    Case 9, 4, 6, 11
    DaysInAnyMonth = 30
    Case 2
    if (IntYear Mod 4 = 0) And (IntYear Mod 100 <> 0) Or (IntYear Mod 400 = 0) Then
    DaysInAnyMonth = 29
    Else
    DaysInAnyMonth = 28
    End if
    Case Else
    DaysInAnyMonth = 31
    End Select
    End function

 
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Other User Comments
3/29/2001 2:09:22 AMJeffgueldre@hotmail.com

I propose an alternative code :
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4/17/2001 11:03:14 AMJeff

DaysInAnyMonth=dateserial(IntYear, intMonth+1,0)

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9/3/2001 10:09:46 AMdcoragem

Remenber, day 0 of the Month is the last day of the previous month.

DaysInAnyMonth = Day(TheYear, TheMonth + 1, 0)

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9/3/2001 3:36:37 PMNery

DaysInAnyMonth = Day(DateSerial(Year, Month + 1, 0))
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10/17/2002 12:35:58 AMCharles Chadwick

Day(DateAdd("d", -1, currentMonth + 1 & "/01/" & currentYear))
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