Notice: This website is an unofficial Microsoft Knowledge Base (hereinafter KB) archive and is intended to provide a reliable access to deleted content from Microsoft KB. All KB articles are owned by Microsoft Corporation. Read full disclaimer for more details.

ACC2000: "W" Option of the DateDiff() Function Does Not Work


View products that this article applies to.

This article was previously published under Q207795
Moderate: Requires basic macro, coding, and interoperability skills.

This article applies to a Microsoft Access database (.mdb) and to a Microsoft Access project (.adp).

For a Microsoft Access 2002 version of this article, see 288194 (http://support.microsoft.com/kb/288194/EN-US/ ) .

↑ Back to the top


Symptoms

When you use the "w" Weekday option to calculate the number of weekdays between two dates, the DateDiff() function returns the number of weeks, not the number of work days. The "w" option is supposed to work the same way as "d" for DateDiff(). It is provided as an option for compatibility with the DatePart() function.

↑ Back to the top


Resolution

If you are using the DateDiff() function to return the number of days, substitute "d" for "w". You can use the Visual Basic code in this article to return the number of work days rather than the number of days.

Microsoft provides programming examples for illustration only, without warranty either expressed or implied. This includes, but is not limited to, the implied warranties of merchantability or fitness for a particular purpose. This article assumes that you are familiar with the programming language that is being demonstrated and with the tools that are used to create and to debug procedures. Microsoft support engineers can help explain the functionality of a particular procedure, but they will not modify these examples to provide added functionality or construct procedures to meet your specific requirements. The following code provides a function, DateDiffW(), that calculates the number of work days between two dates:
Function DateDiffW(BegDate, EndDate)
   Const SUNDAY = 1
   Const SATURDAY = 7
   Dim NumWeeks As Integer

   If BegDate > EndDate Then
      DateDiffW= 0
   Else
      Select Case Weekday(BegDate)
         Case SUNDAY : BegDate = BegDate + 1
         Case SATURDAY : BegDate = BegDate + 2
      End Select
      Select Case Weekday(EndDate)
         Case SUNDAY : EndDate = EndDate - 2
         Case SATURDAY : EndDate = EndDate - 1
      End Select
      NumWeeks = DateDiff("ww", BegDate, EndDate)
      DateDiffW= NumWeeks * 5 + Weekday(EndDate) - Weekday(BegDate)
   End If
End Function
				

How to Use the DateDiffW() Function

Use the DateDiffW() function wherever you would use DateDiff(). Instead of
DateDiff("W",[StartDate],[EndDate])
				
use the following:
DateDiffW([StartDate],[EndDate])
				
NOTE: This function returns the days UP TO the ending date, not UP TO and INCLUDING the ending date.

Steps to Test the DateDiffW() Function

In the Immediate Window, type the following line, and then press ENTER:
?DateDiffW(#2/1/99#,#2/15/99#)
				
Note that 10 is returned, the number of work days.

↑ Back to the top


More information

Steps to Reproduce Behavior

In the Immediate Window, type the following line, and then press ENTER:
? DateDiff("W",#2/1/99#,#2/15/99#)
				
Note that 2 is returned (the number of weeks), not 14 (the number of days) or 10 (the number of work days).

↑ Back to the top


References

For more information about the DateDiff function, click Microsoft Access Help on the Help menu, type DateDiff Function in the Office Assistant or the Answer Wizard, and then click Search to view the topic.

↑ Back to the top


Keywords: KB207795, kbprb, kbdta, kbprogramming

↑ Back to the top

Article Info
Article ID : 207795
Revision : 2
Created on : 6/23/2005
Published on : 6/23/2005
Exists online : False
Views : 280