C#中TreeView完成合適兩級節點的選中節點辦法

C#中TreeView完成合適兩級節點的選中節點辦法。本站提示廣大學習愛好者：（C#中TreeView完成合適兩級節點的選中節點辦法）文章只能為提供參考，不一定能成為您想要的結果。以下是C#中TreeView完成合適兩級節點的選中節點辦法正文

本文實例講述了C#中TreeView完成合適兩級節點的選中節點辦法。分享給年夜家供年夜家參考。詳細以下：

class TreeViewChecked
{
  bool isfirst = true;
  public TreeViewChecked(TreeView treeView)
  {
   treeView.AfterCheck += new TreeViewEventHandler(treeView_AfterCheck);
   treeView.AfterSelect += new TreeViewEventHandler(treeView_AfterSelect);
  }
  /// <summary>
  /// 點擊文字，選中checkbox
  /// </summary>
  /// <param name="sender"></param>
  /// <param name="e"></param>
  void treeView_AfterSelect(object sender, TreeViewEventArgs e)
  {
   if (e.Node.Checked)
    e.Node.Checked = false;
   else
    e.Node.Checked = true;
   if (isfirst)
   {
    isfirst = false;
    e.Node.TreeView.Nodes[0].Checked = false;
   }
  }
  /// <summary>
  /// 選中子節點，勾選父節點，選中父節點，勾選子節點
  /// </summary>
  /// <param name="sender"></param>
  /// <param name="e"></param>
  void treeView_AfterCheck(object sender, TreeViewEventArgs e)
  {
   CheckTreeNode(e.Node);
  }
  /// <summary>
  /// 選中子節點，勾選父節點，選中父節點，勾選子節點
  /// </summary>
  /// <param name="node"></param>
  private void CheckTreeNode(TreeNode node)
  {
   node.TreeView.AfterCheck -= new TreeViewEventHandler(treeView_AfterCheck);
   //假如不等於null,解釋選擇的是子節點
   if (node.Parent != null)
   {
    TreeNode parent = node.Parent as TreeNode;
    //假如該節點是選中的
    if (node.Checked)
    {
     //斷定其父節點能否被選中，假如沒有被選中則選中它
     if (parent.Checked == false)
     {
      parent.Checked = true;
     }
    }
    else
    {
     bool ischecked = false;
     foreach (TreeNode child in parent.Nodes)
     {
      if (child.Checked)
      {
       ischecked = true;
       break;
      }
     }
     if (ischecked)
      parent.Checked = true;
     else
      parent.Checked = false;
    }
   }//假如等於null，解釋選擇的是根節點
   else
   {
    foreach (TreeNode child in node.Nodes)
    {
     child.Checked = node.Checked;
    }
   }
   node.TreeView.AfterCheck += new TreeViewEventHandler(treeView_AfterCheck);
  }
}

願望本文所述對年夜家的C#法式設計有所贊助。