DevExpress实现TreeList节点互斥的方法


本文实例讲述了DevExpress实现TreeList节点互斥的方法,具体实现方法如下所示:

主要功能代码如下:

/// <summary>
/// 节点互斥同步
/// 说明
/// eg:
///TreeListNode _node = e.Node;
///_node.SyncMutexNodeCheckState(_node.CheckState, n => n.GetNodeType() == NodeType.Cab);
/// </summary>
/// <param name="node">需要互斥同步的节点</param>
/// <param name="checkState">节点状态</param>
/// <param name="checkHanlder">互斥条件【委托】</param>
public static void SyncMutexNodeCheckState(this TreeListNode node, CheckState checkState, Predicate<TreeListNode> checkHanlder)
{
  TreeList _tree = node.TreeList;
  if (checkHanlder(node))//当前节点符合互斥条件时候
  {
 _tree.DownRecursiveTree(n => n.CheckState = CheckState.Unchecked);
  }
  else
  {
 TreeListNode _curParentNode = node.GetParentNode(checkHanlder);//获取符合互斥条件的父节点
 if (_curParentNode == null) return;
 TreeListNode _thePubleNode = node.GetPublicParentNode(checkHanlder);//获取符合互斥条件的公共父节点
 if (_thePubleNode == null) return;
 foreach (TreeListNode n in _thePubleNode.Nodes)
 {
   foreach (TreeListNode nc in n.Nodes)
   {
 if (nc != _curParentNode)
 {
   nc.CheckState = CheckState.Unchecked;
   nc.DownRecursiveNode(nr => nr.CheckState = CheckState.Unchecked);
 }
   }
 }
  }
  node.SyncNodeCheckState(checkState);
  node.CheckState = checkState;
}
}
/// <summary>
/// 向上递归,获取符合条件的节点的公共父节点
/// </summary>
/// <param name="node">操作节点</param>
/// <param name="checkHanlder">委托</param>
/// <returns>符合条件的节点</returns>
public static TreeListNode GetPublicParentNode(this TreeListNode node, Predicate<TreeListNode> checkHanlder)
{
  TreeListNode _publicPNode = null;
  TreeListNode _findNode = node.GetParentNode(checkHanlder);//先获取到条件判断的自身父节点
  if (_findNode != null)
  {
 //开始向上递归
 UpwardRecursiveNode(_findNode, n =>
 {
   TreeListNode _curpublicNode = n.ParentNode;//获取当前向上递归的父节点
   if (_curpublicNode != null)
   {
 if (_curpublicNode.Nodes.Count > 1)//若有多个子节点,则是公共父节点
 {
   _publicPNode = _curpublicNode;
   return false;//跳出递归
 }
   }
   return true;//继续递归
 });
  }
  return _publicPNode;
}
/// <summary>
/// 向上递归,获取符合条件的父节点
/// </summary>
/// <param name="node">需要向上递归的节点</param>
/// <param name="conditionHanlder">判断条件【委托】</param>
/// <returns>符合条件的节点【TreeListNode】</returns>
public static TreeListNode GetParentNode(this TreeListNode node, Predicate<TreeListNode> conditionHanlder)
{
  TreeListNode _parentNode = node.ParentNode;//获取上一级父节点
  TreeListNode _conditonNode = null;
  if (_parentNode != null)
  {
 if (conditionHanlder(_parentNode))//判断上一级父节点是否符合要求
 {
   _conditonNode = _parentNode;
 }
 if (_conditonNode == null)//若没有找到符合要求的节点,递归继续
   _conditonNode = GetParentNode(_parentNode, conditionHanlder);
  }
  return _conditonNode;
}

SyncNodeCheckState代码可以参考:http://www.phpstudy.net/article/53335.htm

说明:

如上图所示,节点“Test3”和“蒙自路Test2”都是"cab"类型;
当调用代码如下:

TreeListNode _node = e.Node;
_node.SyncMutexNodeCheckState(_node.CheckState, n => n.GetNodeType() == NodeType.Cab);

实现的效果就是要么只能勾选“Test3”或者“蒙自路Test2”节点或者子节点,不同同时勾选,应该就是互斥的意思;也是这段代码想实现的效果,希望对大家的项目开发有所帮助。


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