Thanks for the info.
I think then we can not use the .Net as tool for our programm
So slow code, this is a very big risk....
Here is the code just for information:
private void AddNode(XmlNode inXmlNode, TreeNode inTreeNode)
{
XmlNode xNode;
TreeNode tNode;
XmlNodeList nodeList;
int i;
// Loop through the XML nodes until the leaf is reached.
// Add the nodes to the TreeView during the looping process.
if (inXmlNode.HasChildNodes)
{
nodeList = inXmlNode.ChildNodes;
for (i = 0; i <= nodeList.Count - 1; i++)
{
xNode = inXmlNode.ChildNodes;
inTreeNode.Nodes.Add(new TreeNode(xNode.Name));
tNode = inTreeNode.Nodes;
AddNode(xNode, tNode);
}
}
else
{
// Here you need to pull the data from the XmlNode based on the
// type of node, whether attribute values are required, and so forth.
inTreeNode.Text = (inXmlNode.OuterXml).Trim();
}
}
private void loadFileToolStripMenuItem_Click(object sender, EventArgs e)
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.InitialDirectory = "c:\\";
openFileDialog1.Filter = "xml files (*.xml)|*.xml|All files (*.*)|*.*";
openFileDialog1.FilterIndex = 1;
openFileDialog1.RestoreDirectory = true;
System.IO.Stream myStream = null;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
try
{
treeView1.Nodes.Clear();
if ((myStream = openFileDialog1.OpenFile()) != null)
{
using (myStream)
{
try
{
DateTime BeginDownload = DateTime.Now;
// SECTION 1. Create a DOM Document and load the XML data into it.
XmlDocument dom = new XmlDocument();
dom.Load(myStream);
// SECTION 2. Initialize the TreeView control.
treeView1.Nodes.Clear();
DateTime now = DateTime.Now;
TimeSpan diff = now - BeginDownload;
this.Text += (" 1=" + diff.TotalSeconds.ToString());
treeView1.Nodes.Add(new TreeNode(dom.DocumentElement.Name));
TreeNode tNode = new TreeNode();
tNode = treeView1.Nodes[0];
// SECTION 3. Populate the TreeView with the DOM nodes.
AddNode(dom.DocumentElement, tNode);
treeView1.ExpandAll();
now = DateTime.Now;
diff = now - BeginDownload;
this.Text += (" 2=" +diff.TotalSeconds.ToString());
}
catch (XmlException xmlEx)
{
MessageBox.Show(xmlEx.Message);
}
catch (Exception ex)
{
MessageBox.Show(ex.Message);
}
}
}
}
catch (Exception ex)
{
MessageBox.Show("Error: Could not read file from disk. Original error: " +
ex.Message);
}
}
}
