Reputation: 1147
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>
Upvotes: 0
Views: 6508
Reputation: 96840
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();
}
}
Upvotes: 0
Reputation: 5929
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).
Upvotes: 0
Reputation: 4563
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
Upvotes: 1