open file & mycomputer

I hope someone can help me, i have made a contextmenu and i want it to goto the mycomputer folder and then when a file is selected that it starts the file now i already have the openfiledialog and such and i use the code listed below. But it will not start in the mycomputer directory and it will not open the file if i change the > to < then it won't cancel if i choose not to open a file can anyone help me get the code fixed ?


PrivateSub ComputerToolStripMenuItem_Click(ByVal senderAs System.Object,ByVal eAs System.EventArgs)Handles ComputerToolStripMenuItem.Click
Dim mycomputerAsString

mycomputer = System.Environment.GetFolderPath( _
Environment.SpecialFolder.MyComputer)
Dim pAsNew System.Diagnostics.Process()
Me.OpenFileDialog1.Title = "My Computer - Open File"
Me.OpenFileDialog1.ShowDialog()
Me.OpenFileDialog1.InitialDirectory = mycomputer

IfMe.OpenFileDialog1.FileName >NothingThen
p.StartInfo.FileName =Me.OpenFileDialog1.FileName
Else
ExitSub

p.Start()
EndIf
EndSub

also i have tried the following code but then when i decide not to open a file it hangs what am i missing that will make it so when i decide not to open anything that it will still continue to run the application and not hang and how do i make it start in the mycomputer folder ?

Dim iAsInteger
OpenFileDialog1.Filter = "All Files (*.*)|*.*"
OpenFileDialog1.RestoreDirectory =True
OpenFileDialog1.ShowDialog()

If OpenFileDialog1.FileName = ""Then

EndIf
Process.Start(OpenFileDialog1.FileName)
ExitSub

I hope someone can help me correct the code so it works.

[3151 byte] By [CyberOps] at [2007-12-16]
# 1
You want to check the openFileDialog's DialogResult to make sure the user clicked Open before passing the .FileName to Process.Start().

Dim mycomputer As String

mycomputer = System.Environment.GetFolderPath( _

Environment.SpecialFolder.MyComputer)


Dim
p As New System.Diagnostics.Process()

Dim dr As New DialogResult()


Me
.OpenFileDialog1.Title = "My Computer - Open File"

Me.OpenFileDialog1.InitialDirectory = mycomputer


dr = Me.OpenFileDialog1.ShowDialog(Me)

If dr = System.Windows.Forms.DialogResult.OK Then

p.StartInfo.FileName = Me.OpenFileDialog1.FileName

p.Start()

End If

BenWillett at 2007-9-9 > top of Msdn Tech,Windows Forms,Windows Forms General...