如何用程序打开txt 文件

小弟不才～刚学第二天～献丑～

Public Class Form1

    Dim File As String

    Private Sub Button2_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles Button2.Click
        Dim FileNum As Integer
        Dim TempString As String, Temp As String

        TempString = "" : Temp = ""

        FileNum = FreeFile()
        If File <> "" Then
            FileOpen(FileNum, File, OpenMode.Input, OpenAccess.Default, OpenShare.Default)
            Do While Not EOF(FileNum)
                Temp = LineInput(FileNum)
                TempString = TempString & Temp & vbCrLf
            Loop
            FileClose(FileNum)
            TextBox2.Text = TempString
        End If
    End Sub

    Private Sub Button1_MouseDown(ByVal sender As Object, ByVal e As System.Windows.Forms.MouseEventArgs) Handles Button1.MouseDown
        Dim FileName As String

        Dialog1.FileName = ""
        Dialog1.Filter = "文字档​​（*. TXT）|*. TXT|所有档案 (*.*)|*.*"
        Dialog1.Title = "Select a TXT File"
        Dialog1.ShowDialog()

        FileName = Dialog1.FileName

        If FileName <> "" Then
            File = FileName
            TextBox1.Text = FileName
        End If

        FileName = "" : TextBox2.Text = ""
    End Sub

End Class

这是可以用～只是我顺便想问～怎么开好慢～是因为和VB6的写法不同吗？
还有～
Button1_MouseDown和Button1_Click有什么差别？Button1_Click不能用说～

[ 本帖最后由 wube 于 2011-7-24 12:12 编辑 ]

我所用的是有指定的txt文件相链接的，最方便的直接用shell即可，不过还是谢谢

喔～终于知道你的“打开”是啥意思了～

[ 本帖最后由 wube 于 2011-7-28 08:19 编辑 ]