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Find 3rd working day of the following month for a given date

Posted on 2014-01-09
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Last Modified: 2014-01-10
I need to calculate three (3) working (or week) days following the last day of the month for a given date. For example, if WEDate = 1/3/14, the result should be 2/5/14 (3rd working (or week) day of the following month. (I'm using the following to return the last day of WEDate's month: LDOM: DateSerial(Year([WEDate]),Month([WEDate])+1,0). Any help is greatly appreciated. Thank you.
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Question by:jaguar5554
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14 Comments
 
LVL 26

Expert Comment

by:jerryb30
ID: 39769076
Need we take holidays into account?
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Author Comment

by:jaguar5554
ID: 39769082
Yes -- my apologies for not including that important detail!
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LVL 26

Expert Comment

by:jerryb30
ID: 39769115
US Holidays? Any year? As far I can tell tell, only New Year's Day, Independence Day and Labor day could fall within the first week, but with multiple years, you still have a multitude of possible holidays.
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Author Comment

by:jaguar5554
ID: 39769127
US Holidays and current year; that is, current year of the WEDate. Specifically, I only need to see the 3rd working (or week) day of the month following the WEDate being evaluated. Clear as mud? :-)
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LVL 26

Expert Comment

by:jerryb30
ID: 39769141
Not really. A date in December will go over to the next year.
You might need a table of US Holidays by year. Which could be as few as 3 dates or as many as you have years of data times three.
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Author Comment

by:jaguar5554
ID: 39769277
I see. I'm assuming there is no way for the DateSerial function to specify a day v. weekday?
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LVL 26

Expert Comment

by:jerryb30
ID: 39769634
No.  I have a partial solution, but there is a bug which I have to mull over at the local tavern.
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LVL 50

Accepted Solution

by:
Gustav Brock earned 500 total points
ID: 39770603
You can use the simple looping function below:

datDateFrom = DateSerial(Year(datDate), Month(datDate) + 1, 1)
datDateNext = DateAddWorkdays(datDateFrom, 3)

Public Function DateAddWorkdays(ByVal datDate As Date, Optional ByVal intWorkdays = 1) As Date

' Add intWorkdays to datDate ignoring weekends and holidays.
'
' 2010-10-07, Cactus Data ApS, CPH

    Dim dbs         As DAO.Database
    Dim rst         As DAO.Recordset
    
    Dim intDays     As Integer
    Dim strDate     As String
    Dim strSQL      As String
    
    strDate = Format(datDate, "yyyy\/mm\/dd")
    
    Set dbs = CurrentDb
    strSQL = "Select Top " & intWorkdays & " HolidayDate From tblHoliday Where HolidayDate >= #" & strDate & "#"
    Set rst = dbs.OpenRecordset(strSQL)
    
    While intDays < intWorkdays
        datDate = DateAdd("d", 1, datDate)
        intDays = intDays + 1
        ' Skip weekends.
        If Weekday(datDate, vbMonday) > Weekday(vbFriday, vbMonday) Then
            datDate = DateAdd("d", 1 + 7 - Weekday(datDate, vbMonday), datDate)
        End If
        ' Skip holidays.
        Do
            rst.FindFirst "HolidayDate = #" & Format(datDate, "yyyy\/mm\/dd") & "#"
            datDate = DateAdd("d", Abs(Not rst.NoMatch), datDate)
        Loop Until rst.NoMatch
    Wend
    
    rst.Close
    Set rst = Nothing
    Set dbs = Nothing
    
    DateAddWorkdays = datDate
    
End Function

Open in new window

Of course, adjust table and field name to those of yours.
/gustav
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Author Comment

by:jaguar5554
ID: 39771246
That, gustav, is a beautiful thing! I should visit the tavern more often myself! LOL

I changed datDateNext = DateAddWorkdays(datDateFrom, 3)
to datDateNext = DateAddWorkdays(datDateFrom, 2) and on intial spot check am getting the exact results I'm looking for. All I need to do is update the tblHoliday table at the beginning of the new year. Genius. Thank you so much!
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LVL 50

Expert Comment

by:Gustav Brock
ID: 39771254
Great! You are welcome.

/gustav
0
 
LVL 26

Expert Comment

by:jerryb30
ID: 39771264
Just to give credit-I visited the tavern. Did not come up with a solution. Gustav may or may not have visited a tavern.
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Author Comment

by:jaguar5554
ID: 39771269
Scratch that! I changed it back to datDateNext = DateAddWorkdays(datDateFrom, 3)
(the data I was testing was entered for the wrong year!!). Your original expression works perfectly. Thank you Thank you again. Expert!
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Author Closing Comment

by:jaguar5554
ID: 39771284
I combed the Internet and none of the solutions I found worked as expected. This solution is flawless. Gustav is an expert and generous in sharing his expertise.
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Author Comment

by:jaguar5554
ID: 39771292
jerryb30 -- I hope you had an enjoyable visit at the pub.
gustav -- treat yourself to a visit at the pub.
:-)
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