how to read data from xml

hi experts, i have this xml below:

<response type="categories_listing" generation_date_gmt="Tue, 18 Jun 2013 10:26:28 GMT">
<url>
http://localhost/sml.xml
</url>
<categories site="test">
<category id="1368" adult="N" name="Arte e Artesanato">
<category id="5460" adult="N" name="Artesanato"></category>
<category id="2662" adult="N" name="Esculturas"></category>
<category id="40285" adult="N" name="Ferramentas e Materiais"></category>
<category id="1370" adult="N" name="Gravuras"></category>
<category id="5464" adult="N" name="Livros"></category>
<category id="1369" adult="N" name="Pinturas"></category>
<category id="1371" adult="N" name="Outros"></category>
</category>
<category id="1051" adult="N" name="Celulares e Telefones">
<category id="3813" adult="N" name="Acessórios para Celulares"></category>
<category id="7475" adult="N" name="Cartões de Memória"></category>
<category id="1055" adult="N" name="Celulares"></category>
<category id="5427" adult="N" name="Nextel"></category>
<category id="1058" adult="N" name="Rádio Amador"></category>
<category id="46195" adult="N" name="Telefones"></category>
<category id="7502" adult="N" name="VoIP"></category>
<category id="2908" adult="N" name="Walkie - Talkies"></category>
<category id="1915" adult="N" name="Outros"></category>
</category>
</response>

Open in new window


now, i can't change the xml, so i want to populate an listview with the category and subcategory. how can i do that
LVL 1
rafaelrglAsked:
Who is Participating?
 
Meir RivkinConnect With a Mentor Full stack Software EngineerCommented:
it worked perfectly:screenshot
0
 
Meir RivkinFull stack Software EngineerCommented:
where's the subcategory?
0
 
rafaelrglAuthor Commented:
well, this


<category id="1368" adult="N" name="Arte e Artesanato">
      <category id="5460" adult="N" name="Artesanato"></category>
      <category id="2662" adult="N" name="Esculturas"></category>
      <category id="40285" adult="N" name="Ferramentas e Materiais"></category>
      <category id="1370" adult="N" name="Gravuras"></category>
     <category id="5464" adult="N" name="Livros"></category>
     <category id="1369" adult="N" name="Pinturas"></category>
     <category id="1371" adult="N" name="Outros"></category>
</category>

i formated above so you can see there is category id="1368" and the others inside are subcategorys.
0
Cloud Class® Course: Amazon Web Services - Basic

Are you thinking about creating an Amazon Web Services account for your business? Not sure where to start? In this course you’ll get an overview of the history of AWS and take a tour of their user interface.

 
Meir RivkinFull stack Software EngineerCommented:
since xml contains hierarchy using tree-view control is better approach.
how would u like to display this kind of structure in listview?
0
 
Fernando SotoRetiredCommented:
Hi rafaelrgl;

The following code uses Linq to XML to query the document and build your ListView control.

using System.Xml.Linq;

// Load the XML file from the file system. Change path to the file below
XDocument doc = XDocument.Load("C:/Working Directory/Response.xml");

// Get a reference to the categories node
XElement categoriesElements = doc.Root.Element("categories");

// Query the document for the needed info.
var results = (from d in categoriesElements.Elements("category")
               select new
                          {
                              Parent = d.Attribute("name").Value,
                              Children = (from c in d.Elements("category")
                                          select c.Attribute("name").Value).ToList()
                          }).ToList();

// Build your list view items
foreach (var node in results)
{
    // New parent node
    var lItem = new ListViewItem(node.Parent);
    foreach (var child in node.Children)
    {
        // Add all children nodes
        lItem.SubItems.Add(child);
    }
    // Attach parent node to the list view control
    listView1.Items.Add(lItem);
}

Open in new window

0
 
rafaelrglAuthor Commented:
i mean tree-view, sorry  sedgwick. how this can be accomplish
0
 
Meir RivkinFull stack Software EngineerCommented:
@FernandoSoto
what did u set to the View property of the listview?
there's no correlation between the parent categories and the child categories when u display them in listview in this matter.
0
 
Meir RivkinFull stack Software EngineerCommented:
here in treeview:
        private void populateTreeview()
        {
            WebRequest req = WebRequest.Create("http://www.mercadolivre.com.br/jm/categsXml?as_site_id=MLB&as_max_level=2");
            WebResponse resp = req.GetResponse();
            StreamReader textReader = new StreamReader(resp.GetResponseStream());
            XmlTextReader xmlReader = new XmlTextReader(textReader);

            xmlReader.XmlResolver = null;
            XmlDocument xmlDoc = new XmlDocument();
            xmlDoc.XmlResolver = null;
            xmlDoc.Load(xmlReader);
            treeView1.Nodes.Clear();
            treeView1.Nodes.Add(new TreeNode(xmlDoc.DocumentElement.Name));
            TreeNode tNode = new TreeNode();
            tNode = (TreeNode)treeView1.Nodes[0];
            addTreeNode(xmlDoc.DocumentElement, tNode);
            treeView1.ExpandAll();

        }
        //This function is called recursively until all nodes are loaded
        private void addTreeNode(XmlNode xmlNode, TreeNode treeNode)
        {
            XmlNode xNode;
            TreeNode tNode;
            XmlNodeList xNodeList;
            if (xmlNode.HasChildNodes) //The current node has children
            {
                xNodeList = xmlNode.ChildNodes;
                for (int x = 0; x <= xNodeList.Count - 1; x++)
                //Loop through the child nodes
                {
                    xNode = xmlNode.ChildNodes[x];
                    treeNode.Nodes.Add(new TreeNode(xNode.Name));
                    tNode = treeNode.Nodes[x];
                    addTreeNode(xNode, tNode);
                }
            }
            else //No children, so add the outer xml (trimming off whitespace)
                treeNode.Text = xmlNode.OuterXml.Trim();
        }

Open in new window

0
 
rafaelrglAuthor Commented:
sedgwick, did this code worked for you, did you tested, because for me did not worked. i does not show the content right.

Imports System.Net
Imports System.Xml
Imports System.IO

Partial Class WebUserControl_CategoriasVertical
    Inherits System.Web.UI.UserControl

    Protected Sub Page_Init(sender As Object, e As System.EventArgs) Handles Me.Init
        populateTreeview()

    End Sub
    Private Sub populateTreeview()
        Dim req As WebRequest = WebRequest.Create("http://www.mercadolivre.com.br/jm/categsXml?as_site_id=MLB&as_max_level=2")
        Dim resp As WebResponse = req.GetResponse()
        Dim textReader As New StreamReader(resp.GetResponseStream())
        Dim xmlReader As New XmlTextReader(textReader)

        xmlReader.XmlResolver = Nothing
        Dim xmlDoc As New XmlDocument()
        xmlDoc.XmlResolver = Nothing
        xmlDoc.Load(xmlReader)
        TreeView1.Nodes.Clear()
        TreeView1.Nodes.Add(New TreeNode(xmlDoc.DocumentElement.Name))
        Dim tNode As New TreeNode()
        tNode = DirectCast(TreeView1.Nodes(0), TreeNode)
        addTreeNode(xmlDoc.DocumentElement, tNode)
        TreeView1.ExpandAll()

    End Sub
    'This function is called recursively until all nodes are loaded
    Private Sub addTreeNode(xmlNode As XmlNode, treeNode As TreeNode)
        Dim xNode As XmlNode
        Dim tNode As TreeNode
        Dim xNodeList As XmlNodeList
        If xmlNode.HasChildNodes Then
            'The current node has children
            xNodeList = xmlNode.ChildNodes
            For x As Integer = 0 To xNodeList.Count - 1
                'Loop through the child nodes
                xNode = xmlNode.ChildNodes(x)
                treeNode.ChildNodes.Add(New TreeNode(xNode.Name))
                tNode = treeNode.ChildNodes(x)
                addTreeNode(xNode, tNode)
            Next
        Else
            'No children, so add the outer xml (trimming off whitespace)
            treeNode.Text = xmlNode.OuterXml.Trim()
        End If
    End Sub

Open in new window

0
 
Fernando SotoConnect With a Mentor RetiredCommented:
Hi rafaelrgl;

The following code uses Linq to XML to query the document and build your TreeView control.
The only difference between this solution and the last is the foreach loop.

using System.Xml.Linq;

// Load the XML file from the file system. Change path to the file below
XDocument doc = XDocument.Load("C:/Working Directory/Response.xml");

// Get a reference to the categories node
XElement categoriesElements = doc.Root.Element("categories");

// Query the document for the needed info.
var results = (from d in categoriesElements.Elements("category")
               select new
                          {
                              Parent = d.Attribute("name").Value,
                              Children = (from c in d.Elements("category")
                                          select c.Attribute("name").Value).ToList()
                          }).ToList();

// Build your TreeView Nodes
foreach (var node in results)
{
    // New TreeView Node
    var tvNode = new TreeNode(node.Parent);
    foreach (var child in node.Children)
    {
        // Add all children nodes
        tvNode.Nodes.Add(new TreeNode(child));
    }
    // Attach parent node to the TreeView control
    treeView1.Nodes.Add(tvNode);
}

Open in new window

0
 
rafaelrglAuthor Commented:
i see, you are using the win32 version, i want for asp.net webforms, and i don't want to show the xml file, but the category like this:


Arte e Artesanato
         Artesanato
         Esculturas
         Ferramentas e Materiais
         Gravuras


and i want to be able to have this as an link, so i can build some propertys like put the id on the link, so on.
0
 
rafaelrglAuthor Commented:
hi fernando, i am not use to Linq.

So how do i make this sample to work on asp.net using vb.net
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.