使用treeview，单击treeview节点上的事件

Question

当单击treeview的子节点或父节点时，我正在尝试打开表单：

public partial class Form1 : Form
{
    public Form1()
    {
        InitializeComponent();
    }

    TreeNode head = new TreeNode("HEAD");

    TreeNode member = new TreeNode("MEMBER ");

    TreeNode submember = new TreeNode("SUB-MEMBER");

    private void Form1_Load(object sender, EventArgs e)
    {
        head.Nodes.Add(member);
        member.Nodes.Add(submember);

        treeView1.Nodes.Add(head);
        treeView1.AfterSelect += new TreeViewEventHandler(treeView1_AfterSelect);

    }

    private void treeView1_AfterSelect(object sender, TreeViewEventArgs e)
    {
           if (treeView1.SelectedNode == member)
              {
                  MemberForm mf = new MemberForm();
                  mf.ShowDialog(); 
              }

           if (treeView1.SelectedNode == head)
              {
                  HeadForm hf = new HeadForm();
                  hf.ShowDialog(); 
              }

           if (treeView1.SelectedNode == submember)
              {
                  SubMemberForm sf = new SubMemberForm();  //is this way of checking that which node is clicked efficient???
                  sf.ShowDialog(); 
              }
    }

}

Answer 1

只要您的树视图中只有三个节点，这可能是有效的。但是，这将要求您为添加的每个新节点编写额外的if语句。如果您试图区分节点深度，最好使用Level属性。

private void treeView1_AfterSelect(object sender, TreeViewEventArgs e)
{
       if (treeView1.SelectedNode.Level == 0)
          {
              HeadForm hf = new HeadForm();
              hf.ShowDialog(); 
          }
       else if (treeView1.SelectedNode.Level == 1)
          {
              MemberForm mf = new MemberForm();
              mf.ShowDialog(); 
          }

       else if (treeView1.SelectedNode.Level == 2)
          {
              SubMemberForm sf = new SubMemberForm();
              sf.ShowDialog(); 
          }
}

Answer 2

这可能有点黑客和滥用标签功能，但你可以这样做：

TreeNode Head = new TreeNode("Head");
Head.Tag = typeof(HeadForm);

private void treeView1_AfterSelect(object sender, TreeViewEventArgs e)
{
  Form toOpen = Activator.CreateInstance((Type)treeView1.SelectedNode.Tag) as Form;

  if(toOpen != null)
    toOpen.ShowDialog();
}