為什么不能夠打開

jxyga111 发布于 2008-09-22 11:00， 844 次点击

Function openfile() As String
        OpenFileDialog1.Title = "open dialog"
        '定義文件過濾器的屬性
        OpenFileDialog1.Filter = "rich text file |*.rch |*.txt"
        OpenFileDialog1.ShowReadOnly = True
        If OpenFileDialog1.ShowDialog = DialogResult.OK Then
            openfile = OpenFileDialog1.FileName
        Else
            openfile = ""
        End If
請幫忙解決

2 回复

qlong07282008-09-22 15:20

Open.Filter = "rich text file |*.rch |*.txt"
Open.FilterIndex = 2
Open.ShowDialog()

jxyga1112008-09-22 15:41

謝謝解決了
Function openfile() As String
        OpenFileDialog1.Title = "open dialog"
        '定義文件過濾器的屬性
        OpenFileDialog1.Filter = "*.txt|*.txt"
        OpenFileDialog1.ShowReadOnly = True
        If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then

            openfile = OpenFileDialog1.FileName
        Else
            openfile = ""
        End If