c# 如何读取Xml 获得数据?

Question

<?xml version="1.0" encoding="gb2312"?>
<manifest identifier="SingleSharableResource_MulitipleFileManifest" version="1.1" xmlns="http://www.imsproject.org/xsd/imscp_rootv1p1p2" xmlns:adlcp="http://www.adlnet.org/xsd/adlcp_rootv1p2" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.imsproject.org/xsd/imscp_rootv1p1p2 imscp_rootv1p1p2.xsd
         http://www.imsglobal.org/xsd/imsmd_rootv1p2p1 imsmd_rootv1p2p1.xsd
         http://www.adlnet.org/xsd/adlcp_rootv1p2 adlcp_rootv1p2.xsd">
<metadata/>
<organizations default="hooyes">
<organization identifier="hooyes">
<item xmlns="" identifier="B_hooyes">
<item xmlns="http://www.imsproject.org/xsd/imscp_rootv1p1p2" identifier="B_01_cnn1_1" isvisible="true">
<title>关于个人</title>
<item identifier="S_men" identifierref="R_S_men" isvisible="true">
<title>男人</title>
</item>
<item identifier="S_women" identifierref="R_S_women" isvisible="true">
<title>女人</title>
</item>
</item>
</item>
</organization>
</organizations>
<resources>
<resource identifier="R_S_men" type="webcontent" adlcp:scormtype="sco" href="unit1/index_content.html">
<metadata>
<schema>ADL SCORM</schema>
<schemaversion>1.2</schemaversion>
<adlcp:location>unit1/index_content.html</adlcp:location>
</metadata>
<file href="unit1/index_content.html"/>
<dependency identifierref="R_D_01_Contentcnn1_1_1"/>
</resource>
<resource identifier="R_S_women" type="webcontent" adlcp:scormtype="sco" href="unit2/index_content.html">
<metadata>
<schema>ADL SCORM</schema>
<schemaversion>1.2</schemaversion>
<adlcp:location>unit2/index_content.html</adlcp:location>
</metadata>
<file href="unit2/index_content.html"/>
<dependency identifierref="R_D_01_Contentcnn1_1_2"/>
</resource>

</resources>
</manifest>xml 的特征如上:
如今想通过方法  fx(xml) 
想得到如下格式数据:序号      节点名称                url
1         关于个人                0
2         男人                    unit1/index_content.html
3         女人                    unit2/index_content.html
..... 
高手能否给个实现例子.
  

Answer 1

用DataSet 的Load方法 就可以了 

Answer 2

试过了..不行的哦,主要是有 xmlns ...

Answer 3

XmlDocument   selectNodes

Answer 4

xmlns,不关这个的事，肯定可以的

Answer 5

System.Xml.XmlDocument dom = new System.Xml.XmlDocument();
dom.LoadXml(xml);
foreach (System.Xml.XmlNode node in dom.ChildNodes)
{   
   Response.Write(Get(node)+"<BR>"); 
}
string Get(System.Xml.XmlNode node)
        {
            string str = node.InnerText;
            if (node.HasChildNodes)
            {
                str = "";
                foreach (System.Xml.XmlNode n in node.ChildNodes)
                {
                    str += Get(n);
                }
            }
            return str;
        }

Answer 6

xmlns 只是XML命名空间.
用XMLDocument(存入内存中读取)
或XMLReader(只向前读取)

Answer 7

用XlmDocument或XMLReader
DataSet的Load方法处理有一定的局限性(文档树不能超过三层；根元素不能有属性；根元素不能超过一个，否则多余的根元素及子元素无效)，不能处理复杂的结构

Answer 8

Program.cs类using System;
using System.Collections.Generic;
using System.Text;
using System.Xml;namespace TestConsoleApplication
{
    class Program
    {
        static void Main(string[] args)
        {
            Test3 te = new Test3();
            te.xmlTest();
            Console.Read();
        }        class Test3
        {
            int i = 0;
            XmlDocument myXml = new XmlDocument();
            Xml xml = new Xml();
            public void xmlTest()
            {
                myXml.Load("E:/test.xml");
                XmlNodeList node = myXml.DocumentElement.ChildNodes;
                XmlNode node2 = node.Item(1);
                foreach (XmlNode var in node2.ChildNodes)
                {
                    foreach (XmlNode var1 in var.ChildNodes)
                    {
                        foreach (XmlNode var2 in var1.FirstChild)
                        {
                            switch (var2.Name)
                            {
                                case "title":
                                    xml.title = var2.InnerText;
                                    break;
                                case "item":
                                    xml.name[i] = var2.InnerText;
                                    ++i;
                                    break;
                                default:
                                    break;
                            }
                        }
                    }
                }
                int j = 0;
                node2 = node.Item(2);
                foreach (XmlNode var in node2.ChildNodes)
                {
                    foreach (XmlNode var1 in var.FirstChild)
                    {
                        switch (var1.Name)
                        {
                            case "adlcp:location":
                                xml.url[j] = var1.InnerText;
                                ++j;
                                break;
                            default:
                                break;
                        }
                    }
                }
               
                for (int i = 0; i < xml.url.Length; i++)
                {
                    if (i == 0)
                    {
                        Console.Write("序号\t\t节点名称\t\tURL");
                        Console.WriteLine();
                        Console.Write(i + 1 + "\t\t" + xml.title + "\t\t0");
                    }
                    else
                    {
                        if (xml.name[i - 1] == null)
                        {
                            break;
                        }
                        Console.Write(i + 1 + "\t\t");
                        Console.Write(xml.name[i-1] + "\t\t");
                        Console.Write(xml.url[i-1] + "\t\t");
                    }
                    Console.WriteLine();
                }
            }
        }
    }
}Xml.cs类using System;
using System.Collections.Generic;
using System.Text;namespace TestConsoleApplication
{
    public class Xml
    {
        public string title;
        public string[] name = new string[10];
        public string[] url = new string[10];
    }
}把你的你的发的那个XML文件放到E盘跟目录下..

Answer 9

哎..真是郁闷哦,简单的xml操作谁都会,但是如何处理. xmlns:adlcp 
没有人给出有用的方法.

Answer 10

SCORM 标准中的这个 imsmanifest.xml 看来还真有点意思.

Answer 11

System.Xml.XmlDocument dom = new System.Xml.XmlDocument();
dom.LoadXml(xml);
foreach (System.Xml.XmlNode node in dom.ChildNodes)
{ 
  Console.Write(Get(node)+" <BR>");
}
string Get(System.Xml.XmlNode node)
        {
            string str = node.InnerText;
            if (node.HasChildNodes)
            {
                str = "";
                foreach (System.Xml.XmlNode n in node.ChildNodes)
                {
                    str += Get(n);
                }
            }
            return str;
        }