Visual Basic

Dim dlg As New OpenFileDialog()
If dlg.ShowDialog() = DialogResult.OK
    IO.File.Copy(dlg.FileName, desinationfile)
End If

<br />
     Dim dlg as New OpenFileDialog<br />
     if dlg.ShowDialog() = DialogResult.Ok then<br />
          System.IO.File.Copy(dlg.FileName, "C:\")<br />
     EndIf<br />
     ' Will copy the chosen file to C:\<br />

Dim dlg as New OpenFileDialog<br />
if dlg.ShowDialog() = DialogResult.Ok then<br />
System.IO.File.Copy(dlg.FileName, "C:\a.txt")<br />
EndIf

<br />
Dim FileDialog As New OpenFileDialog<br />
'FileInfo class has <br />
Dim myFile as System.IO.FileInfo<br />
<br />
With FileDialog<br />
if .ShowDialog() = DialogResults.Ok then<br />
'create FileInfo instance where .Filename is the fullpath of source <br />
'file. For example: C:\Example.txt<br />
myFile = new System.IO.FileInfo(.Filename)<br />
<br />
'FileInfo.Name property indicates the file name<br />
'For example: If the source file is C:\Example.txt than the FileInfo.Name<br />
'property is Example.txt<br />
myFile.CopyTo("D:\" & myFile.Name, false)<br />
<br />
'syntax: FileInfo.CopyTo(destfilename as String, forceoverwrite as Boolean)<br />
<br />
End If<br />
End With