从 vb6 中的类模块中打开文件对话框答案

【问题标题】：Opening a File Dialog from within a class module in vb6从 vb6 中的类模块中打开文件对话框
【发布时间】：2011-02-24 00:24:05
【问题描述】：

我想知道如何在 vb6 中使用类模块打开文件对话框。我知道如何在表单中进行操作，但我必须从类模块中打开它。

【问题讨论】：

标签： vb6 fileopendialog

【解决方案1】：

看看下面的 API：

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

以下是它的使用示例：

    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

【讨论】：

【解决方案2】：

是的，您可以调用 API 来引发此对话框。

尽管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 Class 可能应该是 UserControl...，问题就消失了。

【讨论】：