I want to know how can we open a file dialog from withing a class module in vb6. I know how to do in forms, but I have to open it from within a class module.
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2329 次
2 回答
1
是的,您可以调用 API 来引发此对话框。
大多数时候,这种需求源于一个破碎的范式。一个类不应该有一个 UI。当你真正需要这个时,你的类可能应该是一个用户控件......问题就消失了。
于 2010-05-27T02:49:30.670 回答
1
看看下面的 API:
Private Declare Function GetOpenFileName Lib "comdlg32.dll" Alias _
"GetOpenFileNameA" (pOpenfilename As OPENFILENAME) As Long
这是它的使用示例:
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 Function FileOpenDialog()
Dim sFilter As String
OpenFile.lStructSize = Len(OpenFile)
sFilter = "Text Files (*.txt)" & Chr(0) & "*.TXT" & Chr(0)
OpenFile.lpstrFilter = sFilter
OpenFile.nFilterIndex = 1
OpenFile.lpstrFile = String(257, 0)
OpenFile.nMaxFile = Len(OpenFile.lpstrFile) - 1
OpenFile.lpstrFileTitle = OpenFile.lpstrFile
OpenFile.nMaxFileTitle = OpenFile.nMaxFile
OpenFile.lpstrInitialDir = "C:\"
OpenFile.lpstrTitle = "Select File"
OpenFile.flags = 0
lReturn = GetOpenFileName(OpenFile)
End Function
于 2010-05-26T07:26:16.473 回答