Compare and remove unmatched items from list1 and treeview - vb6

Question

I have list1, treeview, list2
list1 has items
treeview1 has items
list2 empty or has items populating over time
Now I want to loop through list1 compare it with list2 and if items do not match then remove the items from treeview1,list1.
The small problem I am having now is list1 ends up keeping 1 unwanted item.
list1 treeview1 and list2 over time has same items populated but list2 sometimes may not contain the same value as list1 this is why I need to loop and match strings. If not found then remove the strings from list1 and treeview1 the same items only.
This code below you can paste it into fresh vb6 project just add 1 treeview1 list1 and list2 and 2 command button
Private Sub Command1_Click()
List1.AddItem "dezzzzz"
TreeView1.Nodes.Add , , , "dezzzzz" & "/" & "General"
List1.AddItem "tammy8123"
TreeView1.Nodes.Add , , , "tammy8123" & "/" & "General"
List1.AddItem "sarah7232"
TreeView1.Nodes.Add , , , "sarah7232" & "/" & "General"
List2.AddItem "tammy8123"
End Sub
Private Sub Command2_Click()
Dim i As Integer, ii As Integer
Dim iii As Integer, demopacket() As String
For ii = List1.ListCount - 1 To 0 Step -1
Dim lstUserspacket() As String
lstUserspacket() = Split(List1.List(ii), "/")
For iii = 1 To TreeView1.Nodes.Count
demopacket() = Split(TreeView1.Nodes(iii).Text, "/")
For i = List2.ListCount - 1 To 0 Step -1
Dim listpacket() As String
listpacket() = Split(List2.List(i), "/")
If listpacket(0) = demopacket(0) Then
Exit For
End If
Next
If i = -1 Then
On Error Resume Next
List1.RemoveItem ii
TreeView1.Nodes.Remove iii
End If
Next
Next
End Sub
Private Sub Form_Load()
Call Command1_Click
End Sub

Answer 1

My first thought was to simply find the issue with the code you provided. My next thought was realizing the code was too complex for a rather simple task. The code could be as simple as:
Private Sub Command2_Click()
Dim i As Integer
For i = List1.ListCount - 1 To 0 Step -1
If Not ExistsInList(List2, List1.List(i)) Then
TreeView1.Nodes.Remove FindInTree(TreeView1, List1.List(i))
List1.RemoveItem FindInList(List1, List1.List(i))
End If
Next
End Sub
Of course, the controls in question do not provide an Exists or Find method. But it is not a hard task to create you own. If you place the following functions in a Utility module then you can use them in other projects, too:
Public Function ExistsInList(ByVal ListBox As ListBox, ByVal Item As String) As Boolean
Dim i As Integer
For i = 0 To ListBox.ListCount - 1
If ListBox.List(i) = Item Then
ExistsInList = True
Exit Function
End If
Next
End Function
Public Function FindInList(ByVal ListBox As ListBox, ByVal Item As String) As Integer
Dim i As Integer
For i = 0 To ListBox.ListCount - 1
If ListBox.List(i) = Item Then
FindInList = i
Exit Function
End If
Next
End Function
Public Function FindInTree(ByVal TreeView As TreeView, ByVal Item As String) As Integer
Dim i As Integer
For i = 1 To TreeView.Nodes.Count
If InStr(1, TreeView.Nodes(i).Text, Item) > 0 Then
FindInTree = i
Exit Function
End If
Next
End Function