VB中打开一个文件并获得这个文件的路径，下一步要处理这个文件。

Question

Private Sub Command1_Click()

CommonDialog1.ShowOpen

Open CommonDialog1.FileName For Output As #1
Text1.Text = CommonDialog1.FileName
Close #1

Dim i As Integer, j As Integer
Set xlApp = CreateObject("Excel.Application")
xlApp.Visible = False
Set xlBook = xlApp.Workbooks.Open(Text1.Text))
Set xlsheet = xlBook.Worksheets(1)

这里只发了一部分我的目的是为了让

Set xlBook = xlApp.Workbooks.Open(Text1.Text)这句话打开刚才所选中的路径，不知道这里面的text1.text该用什么代替？

Answer 1

Private Sub Command1_Click()
CommonDialog1.CancelError = True
CommonDialog1.DialogTitle = "指定要接收的Txt文件"
CommonDialog1.Filter = "文本文件(*.xls)|*.xls|所有文件(*.*)|*.*"
CommonDialog1.InitDir = App.Path
CommonDialog1.ShowOpen
Text1 = CommonDialog1.FileName
Set xlApp = CreateObject("Excel.Application")
xlApp.Visible = True
Set xlBook = xlApp.Workbooks.Open(Text1)
MsgBox "所要打开的Excel文件已经打开，下面将关闭"
xlBook.Save
xlApp.Quit
Set xlApp = Nothing
End
End Sub
'我在Windows XP试验过。更多代码请参阅我的博客:http://hi.baidu.com/zgmg

Answer 2

CommonDialog1.FileName