Question

我正在使用另一个网站上的代码：

How can I model this class in a database?

我在每个客观记录中都有一个名为“Rank”的字段。它告诉我什么位置。例如：

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

该等级告诉我节点的顺序。但我希望获得所有排名：第三位将是：

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

我正在使用LINQ to SQL。我怎么能这样做？

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

Answer 1

我不确定我是否理解你想要的东西，但是递归地遍历树并为对象分配等级非常简单。这是我掀起的一些快速代码：

    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;
    }

使用这个类：

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

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

输出：

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

Answer 2

从以下内容开始：

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++);
    }
}

这将为您提供每个节点的深度（当然您必须将字段Node添加到您的班级）。

Answer 3

到目前为止，最简单的实现方法是在模型或视图模型中。例如，在Node类中，您可以实现以下属性：

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();
    }
}

通常，实现Siblings属性的最简单方法是

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

如果有一组不属于节点层次结构的顶级节点，那么这不起作用，就像这样。你可以通过创建一个没有出现在UI中的根Node对象来假设这一点 - 在这种情况下，你将TreeView绑定到根节点的Children属性，并且像这样实施Rank：

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

如何在treeView中获取节点级别？

3 个答案: