从 vb6 中的类模块中打开文件对话框

Question

我想知道如何从 vb6 中的类模块中打开文件对话框。我知道如何在表单中执行操作，但我必须从类模块中打开它。

Answer 1

看看下面的 API：

Private Declare Function GetOpenFileName Lib "comdlg32.dll" Alias _
    "GetOpenFileNameA" (pOpenfilename As OPENFILENAME) As Long

这是一个使用中的示例：

    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

Answer 2

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。类不应该有 UI。当您真正需要这个时，您的类可能应该是 UserControl...并且问题就消失了。