ms-access - 如何在 vba(access) 中将变量设置为我的表单列表框之一？

Question

我有一个函数，我想根据字符串参数返回不同的列表框。这是函数:

    Here is the function:
Private Function returnList(name As String) As AccessObject
If name = "app" Then
    returnList = Me.Controls("List61")
    'I have also tried the following: 
    'returnList = Me.List61, returnList = Forms![Daily Reports]![List61]
ElseIf name = "lpar" Then
'..several more cases
End If
End Function

每当我尝试调用它时，我都会收到“运行时错误‘91’:未设置对象变量或 With block 变量”。当我使用调试器时，它告诉我对 list61(Me.list61, Me.Controls("List61")) 的引用为空。

有人知道如何解决这个问题吗？任何帮助我将不胜感激。

Answer 1

最重要的是要注意；因为你现在处理的是“对象”而不是“变量”，你必须在对象变量前面加上“Set”这个词。同时将 AccessObject 类型更改为 ListBox。

Private Function returnList(name As String) As ListBox
If name = "app" Then
    Set returnList = Me.Controls("List61")
    'I have also tried the following:
    'returnList = Me.List61, returnList = Forms![Daily Reports]![List61]
ElseIf name = "lpar" Then
'..several more cases
End If
End Function