Find Node in TreeView

hi

as we can get the node in vb6 i need the VB.net

Code:

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'VB6 code
dim Nd as Nde

set Nd=Treeview.nodes("Khalik") ' khalik is the key

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simple i need to find out if the node already created..
if not i need to add node or skip it....

is there away to find node

simply trap the error that you will receive if you try to add a node with a key that already exists and deal with it in the catch block - i.e. flag a message or generate a new unique id and retry etc.

Cheers..

Quote:

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Originally posted by powdir
simply trap the error that you will receive if you try to add a node with a key that already exists and deal with it in the catch block - i.e. flag a message or generate a new unique id and retry etc.

Cheers..

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thanks powdir .. i belive u got it wrong...:mad:

before i add a node to treeview i need to find out if exists i need to do diff operation and if not then add the node...

i know if i loop around the nodes collection i can find that but do we have a method which does the job....?

No there is no intrinsic method to find a node. You'll need to write one.

Here is a previous post on the subject:
http://www.vbforums.com/showthread.p...hreadid=236652

might be of use to some one like me........

well i have written 2 sample function for finding the node
first will return Boolean wheather Node Found or Not.

Second function Return the Node.

Code:

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Function TV_FindNode(ByVal nNode As TreeNodeCollection, ByVal Str As String) As Boolean
        Dim n As TreeNode
        For Each n In nNode
            If n.Text = Str Then
                n.Text = Str
                Return True
                '    Exit For
            End If
        Next
        Return False
    End Function
    Function TV_FindRetNode(ByVal nNode As TreeNodeCollection, ByVal Str As String) As TreeNode
        Dim n As TreeNode
        For Each n In nNode
            If n.Text = Str Then
                n.Text = Str
                Return n
            End If
        Next
    End Function

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