如何在VB.NET中获取XML文件的节点属性名称？

我能够读取XML文件的节点及属性值，但无法获取属性名称。以下为XML示例：

<?xml version="1.0" encoding="UTF-8"?>
<Root>
<body surname="Rott" name="Pig">
<city>London</city>
</body>
<body surname="Lip" name="Jack">
<city>Los Angeles</city>
</body>
<body colorB="White" colorA="Yellow">
<city>Rome</city>
</body>
</Root>

我想要提取出surname、name、colorB、colorA这类属性名称。我尝试用VB.NET编写了如下代码，但找不到类似reader.GetNameAttribute()的方法来获取属性名称，请问是否存在这样的方法？

OpenFileDialog1.Multiselect = False
OpenFileDialog1.ShowDialog()
Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName)
Dim filePath As String = OpenFileDialog1.FileName
Me.Text = filePath
Dim docXML As New XmlDocument
docXML.Load(filePath)
Dim reader As XmlNodeReader = New XmlNodeReader(docXML)

While reader.Read
    Dim lettura = reader.NodeType
    Dim valueA As String
    Dim attr_name As String = "none"
    'attr_name = reader.GetNameAttribute()     here ? get name attribute????

    valueA = reader.Name
    If reader.HasAttributes Then
        Dim conteggio As Integer = reader.AttributeCount
        For x = 0 To conteggio - 1
            MsgBox("Node is: " + valueA + " and has an attribute named: " + attr_name + " = " + reader.GetAttribute(x))
        Next
    End If
End While

方法1：修改现有XmlNodeReader代码 #

你可以使用XmlNodeReader.MoveToAttribute(index)方法定位到指定索引的属性，然后通过reader.Name获取该属性的名称，修改后的代码如下：

OpenFileDialog1.Multiselect = False
OpenFileDialog1.ShowDialog()
Dim fileName As String = Path.GetFileName(OpenFileDialog1.FileName)
Dim filePath As String = OpenFileDialog1.FileName
Me.Text = filePath
Dim docXML As New XmlDocument
docXML.Load(filePath)
Dim reader As XmlNodeReader = New XmlNodeReader(docXML)

While reader.Read
    Dim valueA As String = reader.Name
    If reader.HasAttributes Then
        Dim conteggio As Integer = reader.AttributeCount
        For x = 0 To conteggio - 1
            ' 移动到指定索引的属性
            reader.MoveToAttribute(x)
            ' 获取属性名称和值
            Dim attr_name As String = reader.Name
            Dim attr_value As String = reader.Value
            MsgBox("Node is: " + valueA + " and has an attribute named: " + attr_name + " = " + attr_value)
            ' 移回原节点，避免后续读取出错
            reader.MoveToElement()
        Next
    End If
End While

方法2：直接遍历XmlDocument的节点属性（更简洁） #

如果不需要使用XmlNodeReader，直接通过XmlDocument的节点集合遍历属性会更直观，代码也更易维护：

OpenFileDialog1.Multiselect = False
OpenFileDialog1.ShowDialog()
Dim filePath As String = OpenFileDialog1.FileName
Me.Text = filePath
Dim docXML As New XmlDocument
docXML.Load(filePath)

' 遍历所有body节点
For Each bodyNode As XmlNode In docXML.SelectNodes("//body")
    Dim nodeName As String = bodyNode.Name
    ' 遍历当前节点的所有属性
    For Each attr As XmlAttribute In bodyNode.Attributes
        MsgBox("Node is: " + nodeName + " and has an attribute named: " + attr.Name + " = " + attr.Value)
    Next
Next

内容的提问来源于stack exchange，提问作者Marco Giglio