Ms access 简单的VBA函数来构建地址
我想在MS Access中构建一个函数,查看表中的6个字段(名称、地址、城市、州、邮政编码、国家),并将它们像可识别的地址一样显示在一起。这看起来很简单,但我在尝试检查字段中是否有空字段时遇到了麻烦。例如,如果地址字段中没有有效字符串,我希望函数显示“无效地址”。任何帮助都将不胜感激。一段简单的代码如下所示:Ms access 简单的VBA函数来构建地址,ms-access,vba,Ms Access,Vba,我想在MS Access中构建一个函数,查看表中的6个字段(名称、地址、城市、州、邮政编码、国家),并将它们像可识别的地址一样显示在一起。这看起来很简单,但我在尝试检查字段中是否有空字段时遇到了麻烦。例如,如果地址字段中没有有效字符串,我希望函数显示“无效地址”。任何帮助都将不胜感激。一段简单的代码如下所示: Function FullAddress(Name As String, Address As String, City As String, State As String, ZIP A
Function FullAddress(Name As String, Address As String, City As String, State As String, ZIP As String, Country As String)
If Address = "" Or City = "" Or Country = "" Then
FullAddress = "invalid address"
Else
FullAddress = Name & vbCrLf & Address & vbCrLf & City & ", " & State & " " & vbCrLf & Country
End If
End Function
但是,当任何字段为空时,该函数将导致“#错误”
有没有关于我做错了什么的想法?问题是,当您将这些字段描述为空或空时,它们实际上包含哪些内容。两种可能性是
“”
)Nz
完成该转换
FullAddress(新西兰([名称],“”)、新西兰([地址],“”)等)
试试这个:
Function FullAddress(Name As String, Address As String, City As String, State As String, ZIP As String, Country As String)
If Nz(Address, "") = "" Or Nz(City, "") = "" Or Nz(Country, "") = "" Then
FullAddress = "invalid address"
Else
FullAddress = Name & vbCrLf & Address & vbCrLf & City & ", " & State & " " & vbCrLf & Country
End If
End Function
由于函数声明包含
地址作为字符串
,这意味着地址
在函数体中永远不能为空。