vba调用函数时Byref参数类型不匹配

时间:2014-02-25 18:30:12

标签: vba access-vba

我有一个重命名表列名的方法。

Public Function Rename_Column(tablename As String, oldcolumn As String, newcolumn As String)
  Dim dbs As Database, tdf As TableDef
  Set dbs = CurrentDb

  For Each tdf In dbs.TableDefs
    If tdf.Name = tablename Then
      For Each fld In tdf.Fields
        If fld.Name = oldcolumn Then
          fld.Name = newcolumn
        End If
      Next
    End If
  Next
  dbs.Close
End Function

我正在其他程序中调用此函数

Public Sub querylistboxitems()
  Dim strTableName As String

  strTableName = "Table1"
  Call Rename_Column(strTableName, "old", "New")
End Sub

但它给出了一个错误“Byref参数类型不匹配”

2 个答案:

答案 0 :(得分:21)

还有其他棘手的情况可能会出现此问题。例如,在一行上声明两个(或更多)变量时:

Dim firstSubMenu, secondSubMenu As CommandBarPopup

现在firstSubMenu的类型为Variant,而secondSubMenu的类型为CommandBarPopup。这可能不是您的意图,也可能是上述错误的原因,通过引用它来引用一个需要CommandBarPopup类型的参数的函数。 在这种情况下,解决方案是将它们声明为两行:

Dim firstSubMenu As CommandBarPopup
Dim secondSubMenu As CommandBarPopup

或者,如果你真的想把它挤在一行上,可以这样做:

Dim firstSubMenu As CommandBarPopup, secondSubMenu As CommandBarPopup

请注意,您必须为每个变量重复“As”类型声明。

答案 1 :(得分:2)

因为你没有定义“fld”而失败了。以下内容比通过所有表/字段循环更直接:

Public Function Rename_Column(tablename As String, oldcolumn As String, newcolumn As String)
    Dim dbs As Database
    Dim tdf As TableDef
    Dim fld As Field

    Set dbs = CurrentDb
    Set tdf = dbs.TableDefs(tablename)
    Set fld = tdf.Fields(oldcolumn)
    Set fld = Nothing
    Set tdf = Nothing
    dbs.Close
    Set dbs = Nothing
End Function
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