Access 2010 VBA 类型不匹配(错误 13)或运行时错误 424(需要对象)

Access 2010 VBA Type mismatch (Error 13), or Runtime error 424 (Object Required)

我在 Access 2010 中创建 VBA 代码以执行 2 个操作时遇到问题:

  1. 如果我的 Access Table 中的 3 个地址行是 NULL,或者匹配城市或国家名称 - 将它们导出到单独的 Excel 文件中需要人工审核

  2. 如果不是,则select一个或多个地址值到不同Access中的"Address"列Table.

这只是试用代码,这意味着我只是想展示我可以在输入所有 ElseIf 条件之前相应地导出值。

下面是代码。 "nmad_address_1" 的第一个实例在我 运行 之后突出显示并收到错误。

Public Sub EditFinalOutput()

'set loop variable
Dim i As Long
Dim ExportFile As String


'Open Reference Set
Set db = CurrentDb
Set qs = db.OpenRecordset("SunstarAccountsInWebir_SarahTest")
Set ss = db.OpenRecordset("1042s_FinalOutput_6")

'Set loop for entirety of recordset
For i = 0 To qs.RecordCount - 1
    'ExportFile = "I:\Tax Team\Tax Team\Sean" & Format(i, "00") & ".xlsx"

    If qs.Fields(("nmad_address_1" Is Null Or "nmad_address_1" = "nmad_city" 
Or "nmad_address_1" = "Webir_Country") And ("nmad_address_2" Is Null Or 
"nmad_address_2" = "nmad_city" Or "nmad_address_2" = "Webir_Country") And 
("nmad_address_3" Is Null Or "nmad_address_3" = "nmad_city" Or 
"nmad_address_3" = "Webir_Country")) Then
    DoCmd.TransferSpreadsheet acImport, 10, 
"SunstarAccountsInWebir_Charlie", "I:\Tax Team\Tax Team\Sean", 1


    Else:   qs.Edit
            qs.Fields("box13c_Address") = 9999
            qs.Update
    End If
    qs.MoveNext
Next i

'Close Reference Set
qs.Close
Set qs = Nothing
ss.Close
Set ss = Nothing

End Sub

这个:

If qs.Fields(("nmad_address_1" Is Null Or "nmad_address_1" = "nmad_city" 
Or "nmad_address_1" = "Webir_Country") And ("nmad_address_2" Is Null Or 
"nmad_address_2" = "nmad_city" Or "nmad_address_2" = "Webir_Country") And 
("nmad_address_3" Is Null Or "nmad_address_3" = "nmad_city" Or 
"nmad_address_3" = "Webir_Country")) Then 

是废话。您正在将字符串与 null 和常量字符串相互比较。您实际上并没有比较任何字段值。我什至不知道哪些应该是字段名,哪些应该是字符串。

此外,您在查询中执行 SomeVar Is Null。在 vba 中,您执行 IsNull(SomeVar).

尝试以下操作:

With qs.Fields
    If IsNull(!nmad_address_1) Or !nmad_address_1 = "nmad_city" Or !nmad_address_1 = "Webdir_Country" 'etc...
End With