C# XML 解析器 - 如何获取：innerxml & innerText（xml 属性不是 xml 元素名称）？

Question

我尝试将我的 XML 文件加载到树视图中，但我没有这样做。只有我得到元素名称。我希望我的树视图加载带有 XML 属性。我的问题是如何获取 XML 属性来加载我的树视图？

亲切的问候，

主窗体 这是我初始化 XML 文件的表单。

public Form1()
{
    InitializeComponent();

    // Create an instance of the open file dialog box.
    // This is test purpose only. In production xml files will come from SQL Database.
    OpenFileDialog openFileDialog1 = new OpenFileDialog();

    // Set filter options and filter index.
    openFileDialog1.Title = "Please Choose XML File";
    openFileDialog1.Filter = "XML Files (.xml)|*.xml|All Files (*.*)|*.*";
    openFileDialog1.FilterIndex = 1;// varsayılan olarak jpg uzantıları göster
    openFileDialog1.Multiselect = false;

    // Call the ShowDialog method to show the dialog box.
    openFileDialog1.ShowDialog();

    XmlDataDocument xmldoc = new XmlDataDocument();
    XmlNode xmlnode;
    FileStream fs = new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.Read);
    xmldoc.Load(fs);
    xmlnode = xmldoc.ChildNodes[1];
    treeView1.Nodes.Clear();
    treeView1.Nodes.Add(new TreeNode(xmldoc.DocumentElement.Name));
    TreeNode tNode;
    tNode = treeView1.Nodes[0];
    AddNode(xmlnode, tNode);
}

XML 加载 这是我的 xml 加载子。

        // XML NODE: ADD
        private void AddNode(XmlNode inXmlNode, TreeNode inTreeNode)
        {
            XmlNode xNode;
            TreeNode tNode;
            XmlNodeList nodeList;
            int i = 0;
            if (inXmlNode.HasChildNodes)
            {
                nodeList = inXmlNode.ChildNodes;
                for (i = 0; i <= nodeList.Count - 1; i++)
                {
                    xNode = inXmlNode.ChildNodes[i];
                    inTreeNode.Nodes.Add(new TreeNode(xNode.Name));
                    tNode = inTreeNode.Nodes[i];
                    AddNode(xNode, tNode);
                }
            }
            else
            {
                inTreeNode.Text = inXmlNode.InnerText.ToString();
            }
        }

这是 XML 代码：

<?xml version="1.0" encoding="utf-8"?>
<Menu>
  <AgencyType id="1" name="WATER" Active="Y">
    <AgencyCode id="1" name="FRESH" Active="Y">
      <TypeOfBills id="1" name="INTKON" Active="Y">
        <PaymentType id="1" name="AA" Active="Y"></PaymentType>
        <PaymentType id="2" name="BB" Active="N" /></PaymentType>
        <PaymentType id="3" name="CC" Active="N"></PaymentType>
      </TypeOfBills>
    </AgencyCode>
  </AgencyType>
</Menu>

原始 XML 文件 这是我构建 xml 文件的“XML Notepad 2007”应用程序。

树视图 XML 加载 当我加载我的 xml 数据后，这就是我的树视图在表单上显示的方式。

Answer 1

谢谢。我做到了。这里是。我使用了“XElement”，它的加载速度非常快。

主要：

using System.Xml.Linq; 

public Form1()
{
    InitializeComponent();

    // Create an instance of the open file dialog box.
    // This is test purpose only. In production xml files will come from SQL Database.
    OpenFileDialog openFileDialog1 = new OpenFileDialog();

    // Set filter options and filter index.
    openFileDialog1.Title = "Please Choose XML File";
    openFileDialog1.Filter = "XML Files (.xml)|*.xml|All Files (*.*)|*.*";
    openFileDialog1.FilterIndex = 1;
    openFileDialog1.Multiselect = false;

    // Call the ShowDialog method to show the dialog box.
    openFileDialog1.ShowDialog();

    FileStream fs = new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.Read);

    var data = XElement.Load(openFileDialog1.FileName);
    TreeNode treeNode = treeView1.Nodes.Add("Menu");
    LoadElements(data, treeNode);

    //Clear ListBox items
    ListBoxMain.Items.Clear();

    //Load ListBox First time
    foreach (TreeNode n in treeView1.Nodes)
    {
            ListBoxMain.Items.Add(n.Text);
    }
}

加载元素

private void LoadElements(XElement xElem, TreeNode treeNode)
{
    foreach (XElement element in xElem.Elements())
    {
        if (element.HasElements)
        {
            if (element.FirstAttribute != null)
            {
                TreeNode tempNode = treeNode.Nodes.Add(element.FirstAttribute.Value + "." + element.Attribute("name").Value);
                LoadElements(element, tempNode);
            }
            else
            {
                LoadElements(element, treeNode);
            }
        }
        else
            treeNode.Nodes.Add(element.FirstAttribute.Value + "." + element.Attribute("name").Value);
    }
}

Answer 2

试试这样的

        else
        {
            inTreeNode.Text = inXmlNode.InnerText.ToString();
            //add new code here
            XmlNode inXmlNode = null;
            XmlAttributeCollection attributes = null;
            inXmlNode.Attributes.CopyTo(atttributes);
            foreach(XmlAttribute attribute in attributes)
            {
               string name = attribute.Name;
               string value = attribute.Value;
            }
        }