Recursively adding node to .NET TreeView

Question

I need to create a menu structure taken from a database table that uses the identifier and ParentID and the ranking that is used to determine the order of nodes.

Root(ID 1, ParentID 0, Rank 1)
  - Node(ID 2, ParentID 1, Rank 1)
    - Node(ID 3, ParentID 2, Rank 1)
      - Node(ID 4, ParentID 3, Rank 1)
      - Node(ID 5, ParentID 3, Rank 2)
    - Node(ID 6, ParentID 2, Rank 2)
    - Node(ID 7, ParentID 2, Rank 3)
  - Node(ID 8, ParentID 1, Rank 2)
    - Node(ID 9, ParentID 8, Rank 1)
    - Node(ID 10, ParentID 8, Rank 2)

I tried to create a function to iterate the SQL data and create this tree structure, but I'm not sure how to handle the depth of the add. I can add the first level of nodes by simply checking to see if they have a ParentID, I can add a second level of nodes to the else condition, however I'm not sure how to add the following hierarchy levels.

Iterating through the database:

using (var command = new SqlCommand(_Query, _Connection))
{
    _Connection.Open();                
    var _Reader = command.ExecuteReader();
    while (_Reader.Read())
    {
        CreateNode((int)_Reader["MenuID"], (int)_Reader["ParentID"], (int)_Reader["Rank"], _Reader["English"].ToString());                    
    }
    _Connection.Close();
}

Creating Nodes:

private void CreateNode(int id, int parentID, int rank, string text)
{            
    if(parentID == -1)
    {
        TreeNode _Node = new TreeNode(text, id.ToString());
        Root.Nodes.Add(_Node);
    }

    if (parentID != -1)
    {
        foreach (TreeNode _Node in Root.Nodes)
        {                    
            if (_Node.Value == parentID.ToString())
            {
                _Node.ChildNodes.Add(new TreeNode(text, id.ToString()) { ShowCheckBox = true } );
            }
        }
    }
}

This is currently not sorting nodes by rank

I expect the HTML output to be similar to the following:

<ul id="1">
    <li>A</li>
    <li>
        <ul id="2">
            <li>B</li>
            <li>
                <ul id="3">
                    <li>C</li>
                    <li>
                        <ul id="4">
                            <li>D</li>
                        </ul>
                    </li>
                    <li>
                        <ul id="5">
                            <li>E</li>
                        </ul>
                    </li>
                </ul>                
            </li>
            <li>
                <ul id="6">
                    <li>F</li>
                </ul>
            </li>
            <li>
                <ul id="7">
                    <li>G</li>
                </ul>
            </li>            
        </ul>
    </li>        
    <li>
        <ul id="8">
            <li>H</li>
            <ul>
                <li>
                    <ul id="9">
                        <li>I</li>
                    </ul>
                </li>
                <li>
                    <ul id="10">
                        <li>J</li>
                    </ul>
                </li>                
            </ul>
        </ul>
    </li>
</ul>

http://jsfiddle.net/RjE7H/

Answer 1

, :

Dictionary<int, TreeNode> ParentCache = new Dictionary<int, TreeNode>();
private void CreateNode(int id, int parentID, int rank, string text)
{
    TreeNodeCollection parentNode = root.Nodes;
    if(parentID != 0)
    {
        TreeNode foundParentNode;
        if (!ParentCache.TryGetValue(parentID, out foundParentNode)
            throw new Exception("Given parentID has not been added to the tree yet - " + parentID.ToString());
        parentNode = foundParentNode.ChildNodes;
    }

    TreeNode newNode = new TreeNode(text, id.ToString());
    parentNode.Add(newNode);
    ParentCache.Add(id, newNode);
}

, . , TreeNodeCollections.

, node , , :

    if(parentID != 0)
    {
        TreeNode foundParentNode;
        //Note: I changed the if logic from, "not TryGetValue" to "TryGetValue"
        if (ParentCache.TryGetValue(parentID, out foundParentNode)
            parentNode = foundParentNode.ChildNodes;
    }

Answer 2

. , .

TreeView tv = new TreeView();
private void populateNode()
{

   for(int i=0;i<5;i++)
   {
      var parent = new TreeNode(i,string.Format("Node{0}",i));
      tv.Nodes.Add(parent);
      for(int j=0;j<=3;j++)
      {
        var child = new TreeNode(j,string.Format("childNode{0}",j)
        parent.ChildNodes.Add(child);
        for(int k=0;k<=3;k++)
        {
          var grandchild = new TreeNode(k,string.Format("grandchildNode{0}",k)
          child.ChildNodes.Add(grandchild);

        }
      }
   }

}

Answer 3

• , ( ).
• ?

- ?

class TempTreeNode
{
    public int MenuID { get; set; }
    public int ParentID { get; set; }
    public int Rank { get; set; }
    public string Lang { get; set; }
}

:

        var nodeList = new List<TempTreeNode>();

        using (var command = new SqlCommand(_Query, _Connection))
        {
            _Connection.Open();
            var _Reader = command.ExecuteReader();

            while (_Reader.Read())
            {
                var node = new TempTreeNode()
                {
                    MenuID = (int)_Reader["MenuID"],
                    ParentID = (int)_Reader["ParentID"],
                    Rank = (int)_Reader["Rank"],
                    Lang = _Reader["English"].ToString()
                };
                nodeList.Add(node);
            }
            _Connection.Close();
        }
        // sorting
        nodeList.Sort((a, b) => a.Rank.CompareTo(b.Rank));
        // creation
        CreateNodes(nodeList);

The method of generating nodes ... just guessing what you want, so this is not complete ...

    private List<TreeNode> CreateNodes(List<TempTreeNode> nodes)
    {
        var rootNodes = new List<TreeNode>();
        foreach (var node in nodes)
        {
            if (node.ParentID == -1)
            {
                TreeNode _Node = new TreeNode(node.Lang, node.MenuID.ToString());
                rootNodes.Add(_Node);
            }
            [...] do whatever...
        }
        return rootNodes;
    }