树形结构的Table中获取某一节点的所有叶子节点字符串

递归算法，将叶子节点存入一个空的stringbuilder变量

 1 private void GetLeafIDSb(StringBuilder result, DataTable dt, string childColumnName, string parentColumnName, int Id)
 2         {
 3             DataRow[] temp1 = dt.Select(parentColumnName + "= " + Id + " ");
 4             if (temp1.Length == 0)
 5             {
 6                 result.Append(Id + ",");
 7             }
 8             else
 9             {
10                 foreach (DataRow item2 in temp1)
11                 {
12                     DataRow[] temp2 = dt.Select(parentColumnName + "=" + item2[childColumnName].ToString() + " ");
13                     GetLeafIDSb(result, dt, childColumnName, parentColumnName, int.Parse(item2[childColumnName].ToString()));
14                 }
15             }
16         }

将上方的方法进行包装，返回叶子节点的字符串

1 private string GetLeafIdStr(DataTable dt, string childColumnName, string parentColumnName, int Id)
2         {
3             StringBuilder sb = new StringBuilder();
4             GetLeafIDSb(sb, dt, childColumnName, parentColumnName, Id);
5             sb = sb.Remove(sb.Length - 1, 1);
6             return sb.ToString();
7         }

使用时直接调用下面这个方法即可，会返回指定节点的叶子节点字符串

posted @ 2014-11-27 10:00  nicky0227  阅读(452)  评论(0)  编辑  收藏  举报