The following code is causing random Run-Time Error '35600' Index out of bounds Ok

Good call.  I will put these changes in place and let you know in a couple days 8)

This error will occurs if keepplace is = to 0. That is because listitems cannot begin with a 0. It always begins with 1.
Check to see if keepplace is equal to 0 and trap it if it is.

S

or just make it equal to 1 no?

You may have a point here. If 0 means the first elemnet, then yes, make it keepplace+1, otherwise, trap for 0.

S

If it is possible that your users or (other) application code is removing nodes/rows, you will also experience this error on a situational basis.

You might make your application more resilient by using the key of the node instead of its index.  This is analogous to using a recordset bookmark instead of the absoluteposition property.

Another thing... this is why you delete list/array items from back-to-front when referencing items by index number.

Quote>
 If it is possible that your users or (other) application code is removing nodes/rows, you will also experience this error on a situational basis.

You might make your application more resilient by using the key of the node instead of its index.  This is analogous to using a recordset bookmark instead of the absoluteposition property.
</quote>
Absolutely,  Can you explain how to do this "using the key of the node instead of its index"

-D-

Key is a property of a node and must be unique within a listview or treeview.  It is usually assigned a value when you create the node.  The control hashes the key string value and creates a numeric value that is used in a lookup table.

If you can derive some unique value from the row you are adding, it will make your job easier.

==========================
Example:
    ListView1.ListItems.Add , "Now", "Now"
    ListView1.ListItems.Add , "is", "is"
    ListView1.ListItems.Add , "the", "the"
    ListView1.ListItems.Add , "time", "time"

iterating through the listitems:
Debug.Print listview1.ListItems(x).Key, listview1.ListItems(x).index

Now            1
the            3
time           4
is             2

But how would you find it again and select it:

keepplace = lv.SelectedItem.Index

won't work then.

Dim keepplace_Key As String
keepplace_Key = lv.SelectedItem.Key

Using your exact code snippet above:

Dim keepplace_Key As String
keepplace_Key = lv.SelectedItem.Key

MsgBox keepplace_Key

The message box is empty when an item is selected....  Please advise.

<<Please advise>>

If you don't supply a key to your items, there will be no key value.

The items come from a database.  There is an ID field but it is hidden from the list.  So I will make my decision as such.