Mega Code Archive
File open dialog
<![CDATA[
Imports System
Imports System.Collections
Imports System.ComponentModel
Imports System.Windows.Forms
Imports System.Data
Imports System.Configuration
Imports System.Resources
Imports System.Drawing
Imports System.Drawing.Drawing2D
Imports System.IO
Imports System.Drawing.Printing
Public Class MainClass
Shared Sub Main()
'Declare a OpenFileDialog object
Dim objOpenFileDialog As New OpenFileDialog
'Set the Open dialog properties
With objOpenFileDialog
.Filter = "Text files (*.txt)|*.txt|All files (*.*)|*.*"
.FilterIndex = 1
.Title = "Demo Open File Dialog"
End With
'Show the Open dialog and if the user clicks the Open button,
'load the file
If objOpenFileDialog.ShowDialog = Windows.Forms.DialogResult.OK Then
Dim allText As String
Try
'Read the contents of the file
allText = My.Computer.FileSystem.ReadAllText( _
objOpenFileDialog.FileName)
'Display the file contents in the TextBox
System.Console.WriteLine(allText)
Catch fileException As Exception
Throw fileException
End Try
End If
'Clean up
objOpenFileDialog.Dispose()
objOpenFileDialog = Nothing
End Sub
End Class
]]>