How can I get the node level in a treeView?

Question

I'm using this code from the another web:

How can I model this class in a database?

I have in each objective record a field named "Rank". It tells me what position is. For instance:

Objective "Geometry": Rank1
|_Objective "Squares": Rank1
|_Objective "Circles": Rank2
|_Objective "Triangle": Rank3
  |_Objective "Types": Rank1
Objectve "Algebra": Rank2
Objective "Trigonometry": Rank3

That rank tells me the order of the nodes. But I want to get all the rank: For the third position will be:

Objective "Geometry": Rank1
|_Objective "Squares": Rank1   -> 1.1
|_Objective "Circles": Rank2
|_Objective "Triangle": Rank3
  |_Objective "Types": Rank1   -> 1.3.1
Objectve "Algebra": Rank2
Objective "Trigonometry": Rank3    -> 3

I'm using LINQ to SQL. How can I do that?

    <TreeView Name="treeView1">
        <TreeView.ItemTemplate>
            <HierarchicalDataTemplate DataType="{x:Type data:Objective}" ItemsSource="{Binding Path=Objectives}" >
                <TextBlock Text="{Binding Name}" />
            </HierarchicalDataTemplate>
        </TreeView.ItemTemplate>
    </TreeView>

I suggest you solve this in the model (not in the view / GUI) by some recursive method. — Dänu, Aug 16 '11 at 18:41
For an example of the recursive function that sets the level see: http://stackoverflow.com/questions/6225123/simulating-cte-recursion-in-c — woggles, Aug 16 '11 at 19:45

score 1 · Answer 1 · answered Aug 16 '11 at 18:44

I'm not sure if I understand what you want, but it's pretty straightforward to walk the tree recursively and assign ranks to the objects. Here's some quick code I whipped up:

    public void Test()
    {
        Objective math = Init();
        RankObjective("", math);
        System.Console.ReadLine();
    }

    private void RankObjective(string rank, Objective objective)
    {
        int count = 1;
        if (!String.IsNullOrEmpty(rank))
            Console.WriteLine(objective.Name + ": " + rank);
        foreach (Objective child in objective.Objectives)
        {
            string newRank = String.IsNullOrEmpty(rank) ? count.ToString() : rank + "." + count.ToString();
            RankObjective(newRank, child);
            count++;
        }
    }

    private Objective Init()
    {
        Objective math = new Objective("Math");
        Objective geometry = new Objective("Geometry");
        geometry.Objectives.Add(new Objective("Squares"));
        geometry.Objectives.Add(new Objective("Circles"));
        Objective triangle = new Objective("Triangle");
        triangle.Objectives.Add(new Objective("Types"));
        geometry.Objectives.Add(triangle);
        math.Objectives.Add(geometry);
        math.Objectives.Add(new Objective("Algebra"));
        math.Objectives.Add(new Objective("Trigonometry"));
        return math;
    }

using this class:

public class Objective
{
    public Objective(string name)
    {
        Name = name;
        Objectives = new List<Objective>();
    }

    public string Name { get; set; }
    public List<Objective> Objectives { get; set; }
}

output:

Geometry: 1
Squares: 1.1
Circles: 1.2
Triangle: 1.3
Types: 1.3.1
Algebra: 2
Trigonometry: 3

score 0 · Answer 2 · answered Aug 16 '11 at 18:46

Start with something like:

void main()
{
    foreach(Node root in nodeCollection)
    {
        this.Recursion(root, 1);
    }
}

void Recursion(Node n, int depth)
{
    node.Depth = depth;
    foreach(Node child in n.Children)
    {
        this.Recursion(child, depth++);
    }
}

This will give you a Depth for each node (of course you have to add the field Node to your class).

score 0 · Answer 3 · answered Aug 16 '11 at 19:20

By far the easiest way to implement this is in the model or view model. For instance, in your Node class, you could implement the following properties:

public Collection<Node> Siblings { /* see below */ }

public Collection<Node> Children { get; set; }

public Node Parent { get; set; }

public int Position 
{ 
   get 
   { 
      return (Parent == null) 
         ? 0  // I don't like magic numbers, but I don't want to make this an int? either
         : Siblings.IndexOf(this) + 1; 
   }
}

public string Rank
{
    get
    {
       return (Parent == null)
          ? Position.ToString()
          : Parent.Rank + "." + Position.ToString();
    }
}

Usually, the simplest way to implement a Siblings property is

public Collection<Node> Siblings 
{ 
   get 
   { 
      return (Parent == null)
         ? null
         : Parent.Children;
   }
}

That doesn't work in the case where there is a collection of top-level nodes that's not part of the node hierarchy, like this. You can dummy this up by creating a root Node object that doesn't appear in your UI - in this case, you'd bind the TreeView to the root node's Children property, and implement Rank like this:

public string Rank
{
    get
    {
       if (Parent == null)
       {
          return null;
       }
       if (Parent.Parent == null)
       {
          return Position.ToString();
       }
       return Parent.Rank + "." + Position.ToString();
    }
}

How can I get the node level in a treeView?

3 Answers3

Linked