Vba 访问列表框属性时出现无效的限定符错误

此处lstName是Access表单中列表框的名称

Private Sub cmdUp(lstName As String, SQLName As String, IDName As String, ColumnName As String)
  Dim sText As String
  Dim pText As String
  'check: only proceed if there is a selected item
  If lstName.ItemsSelected.Count = 1 Then
  (...)

调用过程后：

Call cmdUp(lstSchemaName.Name, "eo_ListSchema", "SchemaID", "SchemaName")

我在这一行得到了错误信息：

If lstName.ItemsSelected.Count = 1 Then

错误是：

Invalid Qualifier

所以基本上VBA不能理解这个名称。是一个列表，它应该从中查找选择了多少项。 我发现在VBA中“字符串不是对象，所以字符串变量上没有可以调用的方法”

应该有一个简单的解决办法，但我找不到。如何处理这类问题

非常感谢您的帮助

---

Edgaras

您可以将listbox作为listbox传递，字符串将不具有任何属性，或者可以传递字符串和表单对象，这样可能更安全

Call cmdUp(Me, lstSchemaName.Name, "eo_ListSchema", "SchemaID", "SchemaName")

Private Sub cmdUp(frm as Form, lstName As String, _
     SQLName As String, IDName As String, ColumnName As String)

   Frm(lstName) ...

---

lstName

传入时，它将只是listbox的字符串名称，而不是实际的listbox实例

由于此

lstName

是一个字符串，因此调用

lstName.ItemsSelected

无效

您应该能够将列表框传入：

Private Sub cmdUp(lstBox As ListBox, SQLName As String, IDName As String, ColumnName As String)
...

然后使用以下命令调用此命令：

Call cmdUp(lstSchemaName, "eo_ListSchema", "SchemaID", "SchemaName")

---

您需要在cmdUp过程中将ListBox作为Lsitbox传递

Private Sub cmdUp(lstName As ListBox, SQLName As String, IDName As String, ColumnName As String)

或使用其名称查找适当的控件，如：

Dim myListBox As ListBox = Me.FindControl(lstName)
if myListBox.ItemSelected.Count = 1 Then
...