You can read XML directly into datasets like Dataset.ReadXml(Path)

Sonu.T wrote: about some UserControls in one panel You could have separate UserControls instead of separate panels (Right click on Project in Solution Explorer -> Select Add -> UserControl). That way, each of your "panels" will be separate entities at design time. If you put everything in panels directly, then the main form holding all the panels would be congested with controls from different panels (when designing the UI from Visual Studio). So you could have one panel to the right of the splitter and UserControl derived class for each of your screens. You could then load the appropriate control depending on the node selected in the tree. Something like TreeNode selectedNode = ...; switch(selectedNode.Text) { case "Weekly Processes"; panel.Controls.Clear(); panel.Controls.Add(weeklyProcessesControl); break; ... } where panel is the System.Windows.Forms.Panel to the right of the splitter and weeklyProcessesControl is the usercontrol. Regards Senthil [MVP - Visual C#] _____________________________ My Blog | My Articles | My Flickr | WinMacro

Hi, have a look to my example: I created a windows-application with a form and one 'TreeView'-component named treeView1 in my code. The only code is written within the Load-event of the form: private void Form1_Load(object sender, EventArgs e) { ContextMenu myRootContextMenu = new ContextMenu(); myRootContextMenu.MenuItems.Add("Edit Root"); ContextMenu myChild1ContextMenu = new ContextMenu(); myChild1ContextMenu.MenuItems.Add("Edit Child 1"); ContextMenu myChild2ContextMenu = new ContextMenu(); myChild2ContextMenu.MenuItems.Add("Edit Child 2"); **TreeNode rootTreeNode = new TreeNode("Root");** rootTreeNode.ContextMenu = myRootContextMenu; rootTreeNode.Nodes.Add("First Node"); rootTreeNode.Nodes[0].ContextMenu = myChild1ContextMenu; rootTreeNode.Nodes.Add("Second Node"); rootTreeNode.Nodes[1].ContextMenu = myChild2ContextMenu; treeView1.Nodes.Add(rootTreeNode); } I created one root treenode 'rootTreeNode' TreeNode rootTreeNode = new TreeNode("Root"); and added two child nodes: rootTreeNode.Nodes.Add("First Node"); rootTreeNode.Nodes.Add("Second Node"); After that, I added this treenode to the TreeView 'treeView1' of the Form. treeView1.Nodes.Add(rootTreeNode); Now we come to the context-menues: I created one root context-menu, and gave it a description ContextMenu myRootContextMenu = new ContextMenu(); myRootContextMenu.MenuItems.Add("Edit Root"); and with the property 'ContextMenu' of the rootTreeNode rootTreeNode.ContextMenu = myRootContextMenu; I made the assignment. The context-menues for the child nodes are ContextMenu myChild1ContextMenu = new ContextMenu(); myChild1ContextMenu.MenuItems.Add("Edit Child 1"); ContextMenu myChild2ContextMenu = new ContextMenu(); myChild2ContextMenu.MenuItems.Add("Edit Child 2"); and assigned in the same way. Hope this helps a little bit Regards Erik

Okay - I can see that your code is not working. I think it is because you don't really understand what the GetValues() method is returning. So, I need the answer to the following questions: * What is SQL Server returning? How many rows? How many columns? * What do you want your output to be? *** Developer Day 4 in Reading, England on 2nd December 2006 - Registration Now Open *** Upcoming Scottish Developers events: * Developer Day Scotland: are you interested in speaking or attending? My: Website | Blog | Photos

Sonu.T wrote: as i could not get any reply..ima doing this again Maybe the reason you didn't get any replies the first time was because no one had any ideas, so reposting it probably won't change anything. Except that you have at least gotten one reply this time, this one.;P only two letters away from being an asset

Hi Sonu You need symmetric algorithm for functionality like encrypting password. In symmetric approach single key is used for encrypting strings. You can visit my blog http://aamironline.blogspot.com/2006/08/simple-encryption-and-decryption-with.html[^] where I have created a Cypher class for such purpose. If you are using login control for authentication then you will have to subscribe to Authenticate event of the control. Following sample code shows you how to do authentication with login authenticate event. protected void Login_Authenticate(object sender, AuthenticateEventArgs e) { if (Login.UserName == "Aamir") e.Authenticated = true; else e.Authenticated = false; } M Aamir Maniar aamirOnline.com

Hashtable table=new Hashtable(); foreach (Control c in Page.Controls) { string cType = c.GetType().ToString(); if (table[cType] == null) table[cType] = new ArrayList(); (table[cType] as ArrayList).Add(c); } IEnumerator i = table.GetEnumerator(); while (i.MoveNext()) { ArrayList al = ((DictionaryEntry)i.Current).Value as ArrayList; Object o1 = al[0]; Response.Write(o1.GetType().ToString()); Response.Write(" : "); foreach (Control c in al) Response.Write(c.ID + " "); Response.Write(" "); }

If one array doesn't work, neither will the other. Standard question #1: What do you mean by "not working"? Standard question #2: What error message do you get? --- b { font-weight: normal; }

Thanx Stefan!! this is exactly what i wanted. iam gratefulto you....thanx once again..:-O Sonu

I think you are in a wrong forum and also don't cross post. Best Regards, Apurva Kaushal

I think Page.Controls collection will solve your problems. You can have all information about collection of web controls placed in a web page along with count. Neeraj