VB 2005 code problem

Posted on 2007-11-14
Last Modified: 2013-11-26

I am having a problem with the following function that I brought over from VBA into VB 2005
The error is at the line marked 'XXXX and it reads .Selected is not a member of a ListView

    Function sListSelection() As String
        'returns a string separatted by commas of indices selected in a list box
        Dim lngIndex As Long
        Dim S As String
        Dim Count As Long
        Count = 0
        With frmR.lsbActionR
            For lngIndex = 1 To .Items.Count
                If .Selected(lngIndex - 1) Then 'XXXX
                    If Count = 0 Then
                        S = CStr(lngIndex - 1)
                        S = S & "," & CStr(lngIndex - 1)
                    End If
                    Count = Count + 1
                End If
            Next lngIndex
        End With
        sListSelection = S
    End Function
Question by:murbro
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LVL 43

Accepted Solution

TimCottee earned 500 total points
ID: 20281202
The best way to do this is to use the .SelectedIndices property of the listview which returns a collection containing the selected indices.

Now you could transform that to a string in a similar way:

    Function sListSelection() As String
        'returns a string separatted by commas of indices selected in a list box
        Dim S As String
        With frmR.lsbActionR
            For Each SelectedIndex As Integer In .SelectedIndices
                        S = S & "," & CStr(SelectedIndex)
        End With
        Return S.Substring(1)
    End Function

Or work directly with the collection.
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20281212
try with:
If .Items(IngIndex).Selected Then

but will be better:
Dim Index as Integer
For lngIndex = 1 To .SelectedIndices.Count
                   Index = SelectedIndices(IngIndex)
                    If Count = 0 Then
                        S = CStr(Index - 1)
                        S = S & "," & CStr(Index - 1)
                    End If
                    Count = Count + 1
Next lngIndex

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