Question

Using VBA to edit record and delete records

Asked by: Dan89

I have table (supplier) which simply combines two primary keys from two other tables Contacts and Items, supplier uses the primary key from items as its own primary key. In a certain form there are two list boxes and an option box with 3 radio buttons. The 3 options are, Do nothing, Change to another free item and delete item. The first list box uses a query to show all items that are currently attached to the contact which is in question (as a contact can have more than one different item) from the supplier table.  The other list box runs a similar query that compares the items in the supplier table and those that are in the items table, and displays those that are not in supplier. These are free items and are awaiting to be combined with a contact in the supplier table. When the first and third options are selected the second list box is disabled, when  the second option is selected the second list box is enabled. What I want to happen is the user selects an option from the first list box, then choose an option, should that be option 1, nothing should happen. When option 2 is selected, the itemID from the first list box is replaced by that in the second list box. When the third option is selected the record selected in the first list box should be deleted. These events are triggered by a button click. I am currently using the code shown to achieve this, nothing happens to any records although there is no errors. Any suggestions?

Private Sub btnSaveChange_Click()
Select Case Me!FrameOptions
Case 1
DoCmd.Close
 
Case 2
    If [LstItems].Value <> 0 Then
    
        Dim OldItems As New ADODB.Recordset
    
        OldItems.Open "Supplier", CurrentProject.Connection, _
            CursorType:=adOpenKeyset, LockType:=adLockBatchOptimistic
            
        Do While Not OldItems.EOF
            If OldItems("ItemID") = [LstItems].Value Then
                OldItems("ItemID") = [lstFreeItems].Value
                OldItems.Update
            End If
            OldItems.MoveNext
        Loop
    Else
        MsgBox ("No items are assigned to this contact")
    End If
    DoCmd.Close
Case 3
    Dim Items As New ADODB.Recordset
    
    Items.Open "Supplier", CurrentProject.Connection, _
        CursorType:=adOpenKeyset, LockType:=adLockBatchOptimistic
        
    Do While Not Items.EOF
        If Items("ItemID") = [LstItems].Value Then
            Items.Delete
            Items.Update
        End If
        Items.MoveNext
    Loop
    DoCmd.Close
End Select
End Sub

                                  
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Asked On
2008-02-06 at 07:53:28ID23141569
Tags

Microsoft

,

Access

,

2007

Topic

Access Coding/Macros

Participating Experts
2
Points
250
Comments
6

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Answers

 

by: ee_rleePosted on 2008-02-06 at 08:01:40ID: 20833052

try this

Private Sub btnSaveChange_Click()
Select Case Me!FrameOptions
Case 1
DoCmd.Close
 
Case 2
    If [LstItems].Value <> 0 Then
    
        Dim OldItems As New ADODB.Recordset
    
        OldItems.Open "Supplier", CurrentProject.Connection, _
            CursorType:=adOpenKeyset, LockType:=adLockBatchOptimistic
            
        Do While Not OldItems.EOF
            If OldItems("ItemID") = [LstItems].Value Then
                OldItems.Edit
                OldItems("ItemID") = [lstFreeItems].Value
                OldItems.Update
            End If
            OldItems.MoveNext
        Loop
    Else
        MsgBox ("No items are assigned to this contact")
    End If
    DoCmd.Close
Case 3
    Dim Items As New ADODB.Recordset
    
    Items.Open "Supplier", CurrentProject.Connection, _
        CursorType:=adOpenKeyset, LockType:=adLockBatchOptimistic
        
    Do While Not Items.EOF
        If Items("ItemID") = [LstItems].Value Then
            Items.Delete
        End If
        Items.MoveNext
    Loop
    DoCmd.Close
End Select
End Sub

                                              
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by: Dan89Posted on 2008-02-06 at 08:19:36ID: 20833244

ee

That brings up an error message saying "Method or Data member not found" and highlights the line      .
.Edit
?
Cheers
Dan

 

by: ee_rleePosted on 2008-02-06 at 08:26:24ID: 20833305

just delete rs.edit and try again

 

by: Dan89Posted on 2008-02-06 at 08:38:44ID: 20833410

That doesnt work either, no error message but the original problem remains.

Cheers
Dan

 

by: LSMConsultingPosted on 2008-02-06 at 08:50:27ID: 20833513

What are the "Data Value" for the 3 options in your Option Group? Open your form in Design view, the select the option button or checkbox you're using and make sure that the Data Values for those controls correspond to the values you are using in your Select Case.

Are you sure you're referring to the lstItems listbox correctly? Generally you use Me.lstItems.Column(0) to refer to a specific value in the listbox (note that Columns are zero-based, so Column(0) is the first column). The "value" of your listbox may not be the actual value you expect.

Are your listboxes set to multiselect? If so, you'll have to modify this code.

Easiest way to verify this is to set a breakpoint in your code, then "run" it by clicking your command button. To set a breakpoint, place your cursor in the first or second line of your code and press F9 ... then click the button and you should be thrown into the Editor window. Use the F8 key to step through your code. You can also use the Immediate window to examine the values of the various items ... for example to see the value of the first column in your lstItems listbox type this in the Immediate window and press Enter:

?Me.lstItems.Column(0)

Also: I don't see the need to use Recordsets to do this ... straight SQL would be faster, assuming you don't need a "handle" on the data afterwards.

For the Second Case:

CurrentProject.Connection.Execute "UPDATE Supplier SET ItemID=" & Me.lstFreeItems & " WHERE ItemID=" & Me.lstItems

Third Case:

CurrentProject.Connection.Execute "DELETE * FROM Supplier WHERE ItemID =" & Me.LstItems

 

by: Dan89Posted on 2008-02-06 at 09:11:48ID: 20833722

Perfect solution, used the SQL statements in the end as wouldnt work the other way.
Cheers
Dan

20120131-EE-VQP-002

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