How to get attribute xml from treeview winforms?

LooBi 141 Reputation points
2022-04-15T05:12:04.197+00:00

There is a xml file.

   <?xml version="1.0" encoding="utf-8"?>  
   <group name="Student">  
     <value>A</value>       
     <value>B</value>  
   </group>  

I need to edit the group's name "Student". but the treeview dosen't show attribute.
How can I get the attribute? Thank you for your helping!

The treeview.
193260-image.png

My Code.

 private void button1_Click(object sender, EventArgs e)  
 {  
     OpenFileDialog dlg = new OpenFileDialog();  
     dlg.Filter = "XML Files (*.xml)|*.xml";  
     if (dlg.ShowDialog() == DialogResult.OK)  
     {  
         try  
         {  
             this.Cursor = Cursors.WaitCursor;  
             XmlDocument xDoc = new XmlDocument();  
             xDoc.Load(dlg.FileName);  
             tvLabel.Nodes.Clear();  
             tvLabel.Nodes.Add(new TreeNode(xDoc.DocumentElement.Name));  
             TreeNode tNode = new TreeNode();  
             tNode = (TreeNode)tvLabel.Nodes[0];  
             addTreeNode(xDoc.DocumentElement, tNode);  
             tvLabel.ExpandAll();  
         }  
         catch (Exception ex)  
         {  
             MessageBox.Show(ex.Message);  
         }  
         finally  
         {  
             this.Cursor = Cursors.Default;  
         }  
     }  
 }  
 private void addTreeNode(XmlNode xmlNode, TreeNode treeNode)  
 {  
     XmlNode xNode;  
     TreeNode tNode;  
     XmlNodeList xNodeList;  
     if (xmlNode.HasChildNodes)  
     {  
         xNodeList = xmlNode.ChildNodes;  
         for (int x = 0; x <= xNodeList.Count - 1; x++)  
         {  
             xNode = xmlNode.ChildNodes[x];  
             treeNode.Nodes.Add(new TreeNode(xNode.Name));  
             tNode = treeNode.Nodes[x];  
             addTreeNode(xNode, tNode);  
         }  
     }  
     else  
         treeNode.Text = xmlNode.OuterXml.Trim();  
 }  

---------Edited April, 18----------
Let's say.
I have this kind of XMl file or more complicated one.
so I would like to see all attributes.

   <?xml version="1.0" encoding="utf-8"?>  
   <Order name="Student">  
     <group name="TES" Caption="A_TES" Display="true">  
     <group name="TOR" Caption="B_TOR" Display="true">  
     <group name="TPQ" Caption="D_TPQ" Display="false">  
     <group name="TEX" Caption="A_TEX" Display="true">  
       <item app="TAL" Page="0">ABC</item>  
   <item app="TAL" Page="0">CBD</item>  
   ...  
   ...  
   <Order name="Plan">  
     <package key="Special" Row="0" Col="0">  
       <group key="Special_VB" Row="0" Col="0">  
     <item>bing</item>   
     <item>bx</item>  

Like this picture.
193700-image.png

Windows Forms
Windows Forms
A set of .NET Framework managed libraries for developing graphical user interfaces.
1,865 questions
C#
C#
An object-oriented and type-safe programming language that has its roots in the C family of languages and includes support for component-oriented programming.
10,517 questions
0 comments No comments
{count} votes

Accepted answer
  1. Jack J Jun 24,476 Reputation points Microsoft Vendor
    2022-04-15T07:43:14.927+00:00

    @LooBi , Welcome to Microsoft Q&A, you could try to use XDocument to get the get attribute value from the xml.

    Code:

     public string FileName { get; set; }  
        private void button1_Click(object sender, EventArgs e)  
        {  
            OpenFileDialog dlg = new OpenFileDialog();  
            dlg.Filter = "XML Files (*.xml)|*.xml";  
            if (dlg.ShowDialog() == DialogResult.OK)  
            {  
                try  
                {  
                    this.Cursor = Cursors.WaitCursor;  
                    XmlDocument xDoc = new XmlDocument();  
                    xDoc.Load(dlg.FileName);  
                    FileName= dlg.FileName;  
                    tvLabel.Nodes.Clear();  
                    tvLabel.Nodes.Add(new TreeNode(xDoc.DocumentElement.Name));  
                    TreeNode tNode = new TreeNode();  
                    tNode = (TreeNode)tvLabel.Nodes[0];  
                    addTreeNode(xDoc.DocumentElement, tNode);  
                    tvLabel.ExpandAll();  
                }  
                catch (Exception ex)  
                {  
                    MessageBox.Show(ex.Message);  
                }  
                finally  
                {  
                    this.Cursor = Cursors.Default;  
                }  
            }  
        }  
      
      
        private void addTreeNode(XmlNode xmlNode, TreeNode treeNode)  
        {  
            XmlNode xNode;  
            TreeNode tNode;  
            XmlNodeList xNodeList;  
            if (xmlNode.HasChildNodes)  
            {  
                xNodeList = xmlNode.ChildNodes;  
                for (int x = 0; x <= xNodeList.Count - 1; x++)  
                {  
                    xNode = xmlNode.ChildNodes[x];  
                    string nodetext = xNode.Name+": ";  
                    Console.WriteLine(nodetext);  
                    if(!nodetext.Contains("text"))  
                    {  
                        if (xNode.Attributes.Count>0)  
                        {  
                            for (int i = 0; i < xNode.Attributes.Count; i++)  
                            {  
                                nodetext+=xNode.Attributes[i].Name+" - "+xNode.Attributes[i].Value;  
                            }  
                        }  
                    }  
                      
      
      
                    treeNode.Nodes.Add(new TreeNode(nodetext));  
                    tNode = treeNode.Nodes[x];  
                    addTreeNode(xNode, tNode);  
                }  
            }  
            else  
            {  
                string nodetext = xmlNode.Name+": ";  
                Console.WriteLine(nodetext);  
                if (!nodetext.Contains("text"))  
                {  
                    if (xmlNode.Attributes.Count>0)  
                    {  
                        for (int i = 0; i < xmlNode.Attributes.Count; i++)  
                        {  
                            nodetext+=xmlNode.Attributes[i].Name+" - "+xmlNode.Attributes[i].Value;  
                        }  
                    }  
                }  
                else  
                {  
                    nodetext=xmlNode.InnerText;  
                }  
                treeNode.Text = nodetext;  
            }  
              
        }  
    

    Result:

    193779-image.png

    Best Regards,
    Jack


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