C#

papadimitriou wrote:

that was a very cynical answer.

Quang Tran Minh wrote:

an editor dialog is displayed showing the detailed information of the currently selected node

Quang Tran Minh wrote:

When the user clicks on an item (which can be either project, package or workitem like explained above)...

baranils wrote:

But some oter fiels can not be duplicated

baranils wrote:

but I'm looking for the best way to manage them

public static void FromFile(string path, KorepetytorMenu menu)
{
	XPathDocument doc = new System.Xml.XPath.XPathDocument(path);
	XPathNavigator nav = doc.CreateNavigator();
	if (!nav.MoveToChild("KorepetytorMenu", "")) return;
	if (!nav.MoveToChild("RootNode", "")) return;
	menu.rootNode = KorepetytorMenuNode.FromXmlNavigator(nav);
	ReadNodeChildsRecursive(menu.rootNode, nav);
}

gajatko wrote:

Is it a correct approach?

static void ReadNodeChildsRecursive(KorepetytorMenuNode rootNode, XPathNavigator nav)
{
	XPathNodeIterator iter = nav.SelectChildren(XPathNodeType.Element);
	while (iter.MoveNext()) {
		if (iter.Current.Name == "Node") {
			rootNode.Nodes.Add(KorepetytorMenuNode.FromXmlNavigator(iter.Current));
			ReadNodeChildsRecursive(rootNode.Nodes[rootNode.Nodes.Count - 1], iter.Current);
		}
	}
}

public static KorepetytorMenuNode FromXmlNavigator(XPathNavigator nav)
{
	return new KorepetytorMenuNode() {
		File = nav.GetAttribute("File", "") ?? "",
		Header = nav.GetAttribute("Header", "") ?? "",
		Image = nav.GetAttribute("Image", "") ?? ""
	};
}

string myFileName = @"C:\test\my.xml";
DataTable data = new DataTable();
data.ReadXml(myFileName);