如何循环实现树形目录每个节点有几个只几点

Question

能否帮我写个
表结构
ID PID NAME
1 0 AAA
2 1 BBB
3 1 CCC
4 2 DDD
5 2 EEE
6 3 FFF
7 4 GGG
8 4 HHH
这样的一个游标，我想获取每个节点下面有几个子节点，比如pid为1的有2个，pid为2的有2个，pid为3的1个

Answer 1

 DataTable dtlModel = new DataTable();
        private void CreateData() {
            dtlModel.Columns.Add("id", typeof(Int32));
            dtlModel.Columns.Add("pid", typeof(Int32));
            dtlModel.Columns.Add("name");            for (int i = 1; i < 6; i++) {
                DataRow dr = dtlModel.NewRow();
                dr["id"] = i;
                dr["pid"] = 0;
                dr["name"] = i.ToString();
                dtlModel.Rows.Add(dr);
                for (int j = 1; j < 6; j++) {
                    DataRow dr1 = dtlModel.NewRow();
                    dr1["id"] = int.Parse(i.ToString() + j.ToString());
                    dr1["pid"] = i;
                    dr1["name"] = i + "_" + j;
                    dtlModel.Rows.Add(dr1);
                    for (int k = 1; k < 6; k++) {
                        DataRow dr2 = dtlModel.NewRow();
                        dr2["id"] = int.Parse(i.ToString() + j.ToString() + k.ToString());
                        dr2["pid"] = dr1["id"];
                        dr2["name"] = i + "_" + j + "_" + k;
                        dtlModel.Rows.Add(dr2);
                    }
                }
            }
        }        private void BindTree(DataRow[] rows, TreeNode node) {
            foreach (DataRow r in rows) {
                if (r["pid"] == DBNull.Value)
                    continue;                //如果是根结点，添加到TreeView控件
                if (r["pid"].ToString() == "0") {
                    TreeNode node1 = new TreeNode();
                    node1.Text = string.Format("<input name='hd_{0}' type='hidden' value='{1}'/>{2}", r["id"], r["id"], r["name"].ToString());
                    node1.Value = r["id"].ToString();
                    
                    //node1.Target = "top";
                    TreeView1.Nodes.Add(node1);
                    BindTree(dtlModel.Select("pid=" + r["id"]), node1);
                }
                else if (node != null) {//否则添加到当前结点
                    TreeNode node1 = new TreeNode();
                    node1.Text = string.Format("<input name='hd_{0}' type='hidden' value='{1}'/>{2}", r["id"], r["id"], r["name"].ToString());
                    node1.Value = r["id"].ToString();
                    //node1.Target = "top";
                    node.ChildNodes.Add(node1);
                    BindTree(dtlModel.Select("pid=" + r["id"]), node1);//递归生成子树
                }
            }
        }

Answer 2

node.childnodes()
大概,估计可以获取吧

Answer 3

哥们儿要在程序里 还是SQL里获取

Answer 4

用递归假如你的 表名 对应实体是  table1, 你的第一节 父id=0;你写个查询  对应父id下的子节点。 比如方法：  List<table1> Getsubidbypid(int pid)得到 List<table1>  tblist,然后 count一下。