Treeview control - getting selected node value from recordset

[TR21Mark]

I have downloaded this snippet from MSDN library. I have a table with 3 fields and a query for the table. The fields are

'qryTVNode' (name of query

TVNID (Autonumber PK ID)
TVNName (Treeview Node Name)
TVNparID (field holds the parent ID it is related to)

I have the treeview populated and everything is working great. Now on to the next step. I need to get the value of the TNNparID of the selected node. I have a textbox on this form txtSelNode that I would like this value placed after selection.

In review of the code, strPointerField is the field in the recordset that would hold the value I need. Can someone review and steer me in the right direction?

Thanks to all in advance

Option Compare Database
Option Explicit

Private Sub Form_Load()

'Downloaded from MS MSDN library

Const strTableQueryName = "qryTVNodes"
      Dim db As Database, rst As Recordset
      Set db = CurrentDb
      Set rst = db.OpenRecordset(strTableQueryName, dbOpenDynaset, _
      dbReadOnly)
      AddBranch _
         rst:=rst, _
         strPointerField:="TVNparID", _
         strIDField:="TVNID", _
         strTextField:="TVNName"
      End Sub

      '================= AddBranch Sub Procedure =========================
      '      Recursive Procedure to add branches to TreeView Control
      'Requires:
      '   ActiveX Control:  TreeView Control
      '              Name:  xTree
      'Parameters:
      '               rst:  Self-referencing Recordset containing the data
      '   strPointerField:  Name of field pointing to parent's primary key
      '        strIDField:  Name of parent's primary key field
      '      strTextField:  Name of field containing text to be displayed
      '===================================================================
      Sub AddBranch(rst As Recordset, strPointerField As String, _
                    strIDField As String, strTextField As String, _
                    Optional varReportToID As Variant)
      On Error GoTo errAddBranch
      Dim nodCurrent As Node, objTree As TreeView
      Dim strCriteria As String, strText As String, strKey As String
      Dim nodParent As Node, bk As String
      Set objTree = Me!xTree.Object
      If IsMissing(varReportToID) Then  ' Root Branch.
         strCriteria = strPointerField & " Is Null"
      Else  ' Search for records pointing to parent.
         strCriteria = BuildCriteria(strPointerField, _
         rst.Fields(strPointerField).Type, _
         "=" & varReportToID)
         Set nodParent = objTree.Nodes("a" & varReportToID)
      End If
      ' Find the first
      rst.FindFirst strCriteria
      Do Until rst.NoMatch
         ' Create a string
         strText = rst(strTextField)
         strKey = "a" & rst(strIDField)
         If Not IsMissing(varReportToID) Then  'add new node to the parent
            Set nodCurrent = objTree.Nodes.Add(nodParent, _
            tvwChild, strKey, strText)
         Else    ' Add new node to the root.
            Set nodCurrent = objTree.Nodes.Add(, , strKey, _
            strText)
         End If
         ' Save your place in the recordset so we can pass by ref for
         ' speed.
         bk = rst.Bookmark
         ' Add  to this node.
         AddBranch rst, strPointerField, strIDField, strTextField, _
         rst(strIDField)
         rst.Bookmark = bk     ' Return to last place and continue search.
         rst.FindNext strCriteria   ' Find next
         
      
         
      Loop
exitAddBranch:
      Exit Sub
      '--------------------------Error Trapping --------------------------
errAddBranch:
      MsgBox "Can't add child:  " & Err.Description, vbCritical, _
      "AddBranch Error:"
      Resume exitAddBranch
      End Sub
                    
Private Sub xTree_NodeClick(ByVal Node As Object)

 
    
    
    
End Sub

 

[TR21Mark]

For those interested, I was able to figure it out. The nodSelected.Key was found and then I trimed the first letter off to give me the matching PK ID. Works great

Dim nodSelected As MSComctlLib.Node ' a variable for the currently selected node
    
    Set nodSelected = Me.xTree.SelectedItem ' get the currently selected node
    
    Me.txtSelNode = nodSelected.Key
    Me.txtKeyFix = Right(nodSelected.Key, Len(nodSelected.Key) - 1)

Seems I need to post more often, As soon as I post I usually get somewhere!

 

[SydneyGeek]

How often does laying out a question logically present you with the answer? For me, very...

Denis