为Treenode创建列表

Question

我知道这一定很简单，但我就是不能让它工作…我试图通过比较来自不同行的两列，并相应地将它们作为节点或叶来为extjs树创建一个列表。这是我的样本数据

ListA  ListB  labelName
 NY           Parent1
        NY    Leaf1
 HI           Parent2
 AK           Parent3

这是我的c#终端..。因此，当我匹配NY时，我应该使用Parent1作为节点，Leaf1作为它的叶，而不是HI或AK…但是这样做会抛出所有的数据作为父数据..甚至是树叶。

            SqlCommand cmd = con.CreateCommand();

            comd.CommandText = "SELECT * FROM myTable";
            con.Open();
            SqlDataReader reader = comd.ExecuteReader();
            while (reader.Read())
            {
                City MyData = new City();

                MyData.ListA = reader["ListA"].ToString().Trim();
                MyData.ListB = reader["ListB"].ToString().Trim();
                MyData.labelName = reader["labelName"].ToString().Trim();
                giveData.Add(MyData);
            }

            int count = 1;

            List<TreeNode> myNode = new List<TreeNode>();
            foreach (City MyData in giveData)
            {
                // 1st foreach
                    if (MyData.ListA != "")
                    {

                        TreeNode treeNode = new TreeNode();
                        treeNode.id = count++;
                        treeNode.name = MyData.labelName;
                        treeNode.leaf = false;

                        List<TreeNode> Level1 = new List<TreeNode>();
                        foreach (City labelName  in giveData)
                        {
                            if (labelName.ListA == labelName.ListB)
                            {// 2nd foreach
                                TreeNode node1 = new TreeNode();
                                node1.id = count++;
                                node1.name = labelName.labelName;
                                node1.leaf = true;

                                Level1.Add(node1);
                            }
                        }

                        treeNode.children = Level1;
                        myNode.Add(treeNode);
                }
            }
            return JsonConvert.SerializeObject(myNode);

我是否应该使用数组来存储每条记录并进行比较……我没主意了.我相信有更好的方法来实现这一点……请帮帮忙

Answer 1

假设数据是您声明的方式，并且"parents“将出现在任何叶子之前，下面是我想出的创建树的一种方法：

截取不相关的代码

使用 Dictionary<string, List<TreeNode>>的更新: LINQ

我创建了一个新类TreeNode和一些用于测试的样本数据：

var MyData = new List<City>
                  {
                     new City {ListA = "AK", ListB = "", labelName = "Alaska"},
                     new City {ListA = "HI", ListB = "", labelName = "Hawaii"},
                     new City {ListA = "", ListB = "HI", labelName = "Hawaii Leaf 1"},
                     new City {ListA = "", ListB = "HI", labelName = "Hawaii Leaf 2"},
                     new City {ListA = "NY", ListB = "", labelName = "New York"},
                     new City {ListA = "", ListB = "NY", labelName = "New York Leaf 1"},
                     new City {ListA = "", ListB = "NY", labelName = "New York Leaf 2"}
                  };

这是一个新方法，它基本上创建了两个列表，一个用于父列表，另一个用于叶列表。然后，我循环遍历树叶以找到任何匹配的父代，并将树叶添加到其中：

var index = 0;
var parents = (from p in MyData
               where p.ListB == ""
               select p).ToDictionary(p => p.ListA, p => new TreeNode { id = index++, name = p.labelName, leaf = false });

var leaves = (from l in MyData
              where l.ListA == ""
              group l by l.ListB into stateGroup
              select stateGroup).ToDictionary(g => g.Key, g => g.ToList());

foreach (var leaf in leaves.Where(leaf => parents.ContainsKey(leaf.Key)))
{
    parents[leaf.Key].children =
        leaf.Value.Select(l => new TreeNode {id = index++, name = l.labelName, leaf = true}).ToList();
}

var myNode = parents.Select(p => p.Value).ToList();

return JsonConvert.SerializeObject(myNode);

我认为这应该比使用列表和List.Find()更有效

Answer 2

你最好的选择可能是Linq --我创建了一个又快又脏的VB.Net解决方案，让你朝着正确的方向前进--主要部分是第二个Linq语句……

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
    Dim x As New List(Of City)

    x.Add(New City With {.ListA = "NY", .ListB = "", .Leaf = "Parent1"})
    x.Add(New City With {.ListA = "", .ListB = "NY", .Leaf = "Leaf1"})
    x.Add(New City With {.ListA = "HI", .ListB = "", .Leaf = "Parent2"})
    x.Add(New City With {.ListA = "AK", .ListB = "", .Leaf = "Parent3"})

    tv1.Nodes.AddRange((From y In x Where y.ListA <> "" Select New TreeNode With {
                                                                    .Name = y.ListA,
                                                                    .Text = y.Leaf}).ToArray)

    For Each nd As TreeNode In tv1.Nodes
        Dim Nm As String = nd.Name
        nd.Nodes.AddRange((From y In x Where y.ListB = Nm Select New TreeNode(y.Leaf)).ToArray)
    Next
End Sub

简单地说，您填充了父节点的第一个ste，然后循环遍历所有节点，并使用ListB =节点名称的任何节点填充它们。

希望这能帮助你朝着正确的方向前进

Answer 3

..。我正在尝试用C#编写这段代码，但这段代码应该可以做到：

        int count = 1;

        List<TreeNode> myNode = new List<TreeNode>();
        foreach (City MyData in giveData)
        {
            // 1st foreach
            if (MyData.ListA != "")
            {

                TreeNode treeNode = new TreeNode();
                treeNode.id = count++;
                treeNode.name = MyData.labelName;
                treeNode.leaf = false;

                foreach (City labelName in giveData)
                {
                    if (MyData.ListA == labelName.ListB)
                    {// 2nd foreach
                        TreeNode node1 = new TreeNode();
                        node1.id = count++;
                        node1.name = labelName.labelName;
                        node1.leaf = true;

                        treeNode.Nodes.Add(node1);
                    }
                }

                myNode.Add(treeNode);
            }
        }
        return JsonConvert.SerializeObject(myNode);

希望这至少能让你朝着正确的方向前进。