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How to Delete Selected Items of ListView and ListBox (Snippets)

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20 Mar 2011CPOL 123.2K   7   4
I have some lessons learned on the topic.

With ListView and ListBox, you can have multiple selections when you set the property to do so. There are different ways to delete all selected items from a ListView and a ListBox, right or wrong. I have some lessons learned on the topic.
1. Use For Each loop in ListView. (It works)

VB
For Each i As ListViewItem In ListView1.SelectedItems
    ListView1.Items.Remove(i)
Next

2. Use For Each loop in ListBox. (It doesn’t work)

VB
'For Each i As Object In ListBox1.SelectedItems
'    ListBox1.Items.Remove(i)
'Next
  • The For-Each loop does not work correctly with ListBox as it does with ListView. It causes a System.InvalidOperationException: List that this enumerator is bound to has been modified. An enumerator can only be used if the list does not change.

3. A fix to the problem is to copy the selected items to a list.

VB
Dim lst As New List(Of Object)
For Each a As Object In ListBox1.SelectedItems
    lst.Add(a)
Next
For Each a As Object In lst
    ListBox1.Items.Remove(a)
Next

4. Use Index, the right way and the wrong way.

The following snippet works but it must use an inverse loop.

VB
For i As Integer = ListBox1.SelectedIndices.Count - 1 To 0 Step -1
    ListBox1.Items.RemoveAt(ListBox1.SelectedIndices.Item(i))
Next

The other way (using a normal loop) will not work.

VB
'For i As Integer = 0 To ListBox1.SelectedIndices.Count - 1
'    ListBox1.Items.RemoveAt(ListBox1.SelectedIndices.Item(i))
'Next

But amazingly, the following modification works.

VB
For i As Integer = 0 To ListBox1.SelectedIndices.Count - 1
    ListBox1.Items.RemoveAt(ListBox1.SelectedIndex)
Next

5. Use Do While loop. (This would be the best one based on its clear logic)

VB
Do While (ListBox1.SelectedItems.Count > 0)
    ListBox1.Items.Remove(ListBox1.SelectedItem)
Loop

In fact, the ListBox1.SelectedItem (or SelectedIndex) is always the first one of the ListBox1.SelectedItems (or SelectedIndices).

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


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Comments and Discussions

 
Suggestionother "best" solution? Pin
rpdanjou10-Sep-19 21:43
rpdanjou10-Sep-19 21:43 
QuestionTHX UUUUUUUUUUUUUUUUU Pin
Riccardo Loi16-Aug-15 10:51
Riccardo Loi16-Aug-15 10:51 
QuestionWOW Pin
Member 1044903826-Oct-14 0:46
Member 1044903826-Oct-14 0:46 
GeneralHey, nifty. Thanks. Pin
Br.Bill21-Mar-11 16:20
Br.Bill21-Mar-11 16:20 

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