Get The FileName For The Proces.Start(FileName) Method?

May 27, 2009

I have a Listview on my form with different files. I can selet a file en open it with the proces start method. To open the file i use the OpenFileDialog method and select the filename. This work very good. With the code below.

My question is if there is a method to get the selected file without using the OpenFileDialog Box. So when i dubbelclick the selected file the proces will start with open the selected file. Something like:

proc.StartInfo.FileName = Me.SelectedCell.FileName
instead of proc.StartInfo.FileName = Me.OpenFileDialog1.FileName

PS) May be i can preduce the Filename using the OpenFileDialog method on the background.

Private Sub ListView1_MouseDoubleClick(ByVal sender As Object, ByVal e As System.Windows.Forms.MouseEventArgs) Handles ListView1.MouseDoubleClick
If Me.OpenFileDialog1.ShowDialog() = Windows.Forms.DialogResult.OK Then

[Code]....

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[code].....

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If openFileDialog1.ShowDialog() = DialogResult.OK Then
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[Code].....

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[Code]...

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i use this:

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[code]....

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[Code]...

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[Code]....

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[Code]...

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