有关VB公共对话框的问题
我在VB6.0中写了一段公共对话框的程式，可以显示打开文件的控件，但不能打开所选中的问件，请各位大师指点一下，好吗？
[解决办法]
打开文件要靠其他代码，根据你的文件情况来定
[解决办法]
'文件对话框只能获得所选择文件的路径并不能打开文件。
'测试以下代码，它能打开windows能识别的文件。
Option Explicit
Private Declare Function ShellExecute Lib "shell32.dll" Alias "ShellExecuteA" _
(ByVal hwnd As Long, ByVal lpOperation As String, ByVal lpFile As String, _
ByVal lpParameters As String, ByVal lpDirectory As String, ByVal nShowCmd As Long) As Long
Private Declare Function GetOpenFileName Lib "comdlg32.dll" Alias _
"GetOpenFileNameA" (pOpenfilename As OPENFILENAME) As Long
Private Type OPENFILENAME
lStructSize As Long
hwndOwner As Long
hInstance As Long
lpstrFilter As String
lpstrCustomFilter As String
nMaxCustFilter As Long
nFilterIndex As Long
lpstrFile As String
nMaxFile As Long
lpstrFileTitle As String
nMaxFileTitle As Long
lpstrInitialDir As String
lpstrTitle As String
FLAGS As Long
nFileOffset As Integer
nFileExtension As Integer
lpstrDefExt As String
lCustData As Long
lpfnHook As Long
lpTemplateName As String
End Type
Private Sub Command1_Click()
Dim ofn As OPENFILENAME
Dim rtn As String
ofn.lStructSize = Len(ofn)
ofn.hwndOwner = Me.hwnd
ofn.hInstance = App.hInstance
ofn.lpstrFilter = "TEXT (*.TXT)" & Chr$(0) & "*.TXT" & Chr$(0)
ofn.lpstrFile = Space(254)
ofn.nMaxFile = 255
ofn.lpstrFileTitle = Space(254)
ofn.nMaxFileTitle = 255
ofn.lpstrInitialDir = App.Path
ofn.lpstrTitle = "打开文件"
ofn.FLAGS = 6148
rtn = GetOpenFileName(ofn)
If rtn >= 1 Then
ShellExecute Me.hwnd, "open", ofn.lpstrFileTitle, vbNullString, _
Left(ofn.lpstrFile, InStrRev(ofn.lpstrFile, "") - 1), 1
End If
End Sub
