Listing the last and 15th day of every month in a combo box.

I have issues getting this code work. I have been able to link it to a fill box and generate the dates but only either the 15th of each month is listed  or the last day of the month. Below is the code I have been working on. any help is very much appreciated.

Function ListMondays(fld As Control, id As Variant, _
    row As Variant, col As Variant, code As Variant) _
     As Variant
    Dim intOffset As Integer
    Dim temp As Integer

        temp = 0
   

    Select Case code
        Case acLBInitialize            ' Initialize.
            ListMondays = True
        Case acLBOpen                    ' Open.
            ListMondays = Timer        ' Unique ID.
        Case acLBGetRowCount            ' Get rows.
            ListMondays = 10
        Case acLBGetColumnCount    ' Get columns.
            ListMondays = 1
        Case acLBGetColumnWidth    ' Get column width.
            ListMondays = -1            ' Use default width.
        Case acLBGetValue                ' Get the data.
            'ListMondays = DateSerial(Year(Date), Month(Date) + 1 + row, 1) - 1
        'Case acLBGetFormat
            'ListMondays = "mm/dd/yyyy"
           
            'intOffset = Abs((9 - Weekday(Now)) Mod 7)
           
           Select Case temp
           Case Abs((temp) Mod 2) = 0
                ListMondays = DateSerial(Year(Date), Month(Date) + temp, 1) - 1
                'row = row + 1
                'ListMondays = DateSerial(Year(Date), Month(Date) + temp, 1) + 14
           Case Abs((temp) Mod 2) = 1
               ListMondays = DateSerial(Year(Date), Month(Date) + row, 1) + 14
                           
          End Select
           
            temp = temp + 1
                       
    End Select
SheogorathAsked:
Who is Participating?
 
Gustav BrockCIOCommented:
Jeff and Ray, this is a callback function "selfgenerating" the values. No tables needed.

Here is how to obtain the list:
Public Function ListFifteenthAndLastDayOfMonthInYear( _
  ctl As Control, _
  lngId As Long, _
  lngRow As Long, _
  lngCol As Long, _
  intCode As Integer) As Variant
  
  ' 2011-04-27. Cactus Data ApS, CPH.

  ' Count of months.
  Const cintMonths    As Integer = 12
  ' Fifteenth day of month.
  Const cintFifteenth As Integer = 15
  ' Last day of month.
  Const cintLastDay   As Integer = 0
  ' Format for ListBox.
  Const cstrFormat    As String = "yyyy\-mm\-dd"
  
  Static intYear      As Integer
  Static strFormat    As String
  
  Dim varValue        As Variant
  
  Select Case intCode
    Case acLBInitialize
      intYear = Year(Date)
      If ctl.ControlType = acComboBox Then
        ' Use format of control.
        strFormat = ctl.Format
      Else
        ' Use predefined format.
        strFormat = cstrFormat
      End If
      varValue = True             ' True to initialize.
    Case acLBOpen
      varValue = Timer            ' Autogenerated unique ID.
    Case acLBGetRowCount          ' Get rows.
      varValue = cintMonths * 2   ' Set number of rows.
    Case acLBGetColumnCount       ' Get columns.
      varValue = 1                ' Set number of columns.
    Case acLBGetColumnWidth       ' Get column width.
      varValue = -1               ' Use default width.
    Case acLBGetValue             ' Get the data.
      If lngRow Mod 2 = 0 Then
        ' Get the fifteenth day of this month.
        varValue = DateSerial(intYear, Int(lngRow / 2) + 1, cintFifteenth)
      Else
        ' Get the last day of this month.
        varValue = DateSerial(intYear, -Int(-lngRow / 2) + 1, cintLastDay)
      End If
    Case acLBGetFormat            ' Format the data.
      varValue = strFormat        ' Apply format of the generated strings.
    Case acLBEnd
      ' Do something when the form with the control closes or
      ' the control is requeried.
  End Select
  
  ' Return Value.
  ListFifteenthAndLastDayOfMonthInYear = varValue

End Function

Open in new window

/gustav
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GRayLCommented:
If you want this for say a two year period, you need a small table Nums with an integer field Num containing the values 0-23.  Then run this query:

SELECT DateSerial(Year(Date), Month(Date) + Num, 15) as MidMonthDate,
DateSerial(Year(Date), Month(Date)+Num+1,0) As EndMonthDate FROM Nums;
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GRayLCommented:
If you like what you see, then make that query the Control Source of your Combo Box.
0
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GRayLCommented:
Sorry, that should be Row Source, not Control Source.
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Jeffrey CoachmanMIS LiasonCommented:
Since you did not post any source data, I am not sure how any dates are represented in your DB...

But if you have a table (or query) listing all sequential dates, then something roughly like this will list all the 15ths and all the last days of each month, in a single column, in a combobox:

SELECT DISTINCT DateSerial(Year([YourDateField]),Month([YourDateField])+1,0) AS 15ofMonthOrEOM
FROM YourTable
UNION SELECT IIf(Day([YourDateField])=15,[YourDateField],Null)
FROM YourTable
WHERE (((IIf(Day([YourDateField])=15,[YourDateField],Null)) Is Not Null));

This can probably be simplified, ..but it works as-is...

Sample attached

JeffCoachman
Access-EEQ26974756-List15th-And-.mdb
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SheogorathAuthor Commented:
This is great! Thank you very much :)  ComboBox
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GRayLCommented:
Gustav:  Hi, I appreciate what code can do - I just prefer to minimize the use of code in db development.  I guess I'm a bit quirky that way ;-)
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SheogorathAuthor Commented:
Well I learned three approaches to achieve this. Thanks again to all.
0
 
Gustav BrockCIOCommented:
You are welcome!

Callback functions is one of the gems of Access. Perfect and fast for tasks like this.

/gustav
0
 
Jeffrey CoachmanMIS LiasonCommented:
<Jeff and Ray, this is a callback function "selfgenerating" the values. No tables needed.>

As always...
    "Everyday is a school day"

;-)

Jeff
0
 
GRayLCommented:
I still say a simple table with 24 rows and a small query is better than 60 lines of code in terms of maintenance - in a DB environment.
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Gustav BrockCIOCommented:
But Ray, there is no maintenance for this callback function.
And a simple table wouldn't do it; the last day of February changes between 28 and 29.

/gustav
0
 
GRayLCommented:
Gustav:  This is just a case of individual preference.  BTW, my approach handles leap years.  By maintenance I thinking in four years time after the database designer has moved on the boss decides he wants it on the 14th and second last day of every month - I just change two values in the query.  
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Gustav BrockCIOCommented:
> I just change two values in the query.  

And I just change one!

 ' Fifteenth day of month minus one.
  Const cintFifteenth As Integer = 14

/gustav
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GRayLCommented:
You need more than that!  - you missed the second-last day of the month.  
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Gustav BrockCIOCommented:
OK, but I guess you have figured out how:

  ' Last day of month minus one.
  Const cintLastDay   As Integer = -1

/gustav
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GRayLCommented:
Bingo!  Next thread.
0
 
SheogorathAuthor Commented:
Haha you guys are awesome :D
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