winform怎么解析这段XML

Question

1<?xml version="1.0" encoding="gb2312" ?> 
2<function>
3<function_id>infolist</function_id> 
4<out_code>0</out_code> 
5<out_text>操作成功</out_text> 
6<rows>1802</rows> 
7<DrugInfoList>
8<row>
9<classCode>011405</classCode> 
10<drugType>1</drugType> 
11<drugCode>01140509</drugCode> 
12<drugCode2></drugCode2> 
13<useLevel>2</useLevel> 
14<drugName>硫普罗宁</drugName> 
15 
16</row>
17<row>
18<classCode>011405</classCode> 
19<drugType>1</drugType> 
20<drugCode>01140510</drugCode> 
21<drugCode2></drugCode2> 
22<useLevel>2</useLevel> 
23<drugName>水飞蓟宾</drugName> 
24 
25</row>
26</DrugInfoList>
27</function>
请问怎么解析这段XML谢谢

Answer 1

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Xml.Linq;
using System.Windows.Forms;namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }        private void button1_Click(object sender, EventArgs e)
        {
            string xml = @"
                <function>
                 <function_id>infolist</function_id>  
                  <out_code>0</out_code>  
                  <out_text>操作成功</out_text>  
                  <rows>1802</rows>  
                   <DrugInfoList>
                    <row>
                     <classCode>011405</classCode>  
                     <drugType>1</drugType>  
                     <drugCode>01140509</drugCode>  
                     <drugCode2></drugCode2>  
                     <useLevel>2</useLevel>  
                    <drugName>硫普罗宁</drugName>
                   </row>
                   <row>
                    <classCode>011405</classCode>  
                    <drugType>1</drugType>  
                    <drugCode>01140510</drugCode>  
                    <drugCode2></drugCode2>  
                    <useLevel>2</useLevel>  
                    <drugName>水飞蓟宾</drugName>
                  </row>
                 </DrugInfoList>
                </function>
                ";
            XElement xe = XElement.Parse(xml);
            var out_code = xe.Element("out_code").Value;
            var out_text = xe.Element("out_text").Value;
            var drugNames = xe.Descendants("drugName").Select(x => x.Value);
            MessageBox.Show(out_code);
            MessageBox.Show(out_text);
            foreach (var i in drugNames)
            {
                MessageBox.Show(i);
            }
        }
    }
}

Answer 2

操作XML文件，可以使用XMLDocument对象步骤
一、引入命名空间 System.xml;
二、实例化XmlDocument对象
三、利用XmlDocument对象的Load()方法将指定的XML文件加载入XmlDocument对象
四、取得根节点对象
五、遍历其子节点
  
示例using System.XMl;
Class Program
{
    static void Main()
    {
        XmlDocument myxml = new XmlDocument();
        myxml.Load("student.xml");
        XmlNode rootNode = myxml.DocumentElement;
        foreach(XmlNode detailNode in rootNode.ChildNodes)
        {
                //执行操作
        }
    }
}

Answer 3

string xml = @"
                <function>
                 <function_id>infolist</function_id>  
                  <out_code>0</out_code>  
                  <out_text>操作成功</out_text>  
                  <rows>1802</rows>  
                   <DrugInfoList>
                    <row>
                     <classCode>011405</classCode>  
                     <drugType>1</drugType>  
                     <drugCode>01140509</drugCode>  
                     <drugCode2></drugCode2>  
                     <useLevel>2</useLevel>  
                    <drugName>硫普罗宁</drugName>
                   </row>
                   <row>
                    <classCode>011405</classCode>  
                    <drugType>1</drugType>  
                    <drugCode>01140510</drugCode>  
                    <drugCode2></drugCode2>  
                    <useLevel>2</useLevel>  
                    <drugName>水飞蓟宾</drugName>
                  </row>
                 </DrugInfoList>
                </function>
                ";
            XElement xe = XElement.Parse(xml);
            //获取单个字节点的值:
            string functionId = xe.Element("function_id").Value;
            Console.WriteLine(functionId);
            //遍历DrugInfoList节点下的所有子节点的值
            IEnumerable<XElement> ie= xe.Elements("DrugInfoList").Elements();
            foreach (XElement x in ie)
            {
                Console.WriteLine("NodeName="+x.Name);
                foreach (XElement e in x.Elements())
                Console.WriteLine("Name:"+e.Name+",Value:"+e.Value);
            }

Answer 4

 string xml = @"
                <function>
                 <function_id>infolist</function_id>  
                  <out_code>0</out_code>  
                  <out_text>操作成功</out_text>  
                  <rows>1802</rows>  
                   <DrugInfoList>
                    <row>
                     <classCode>011405</classCode>  
                     <drugType>1</drugType>  
                     <drugCode>01140509</drugCode>  
                     <drugCode2></drugCode2>  
                     <useLevel>2</useLevel>  
                    <drugName>硫普罗宁</drugName>
                   </row>
                   <row>
                    <classCode>011405</classCode>  
                    <drugType>1</drugType>  
                    <drugCode>01140510</drugCode>  
                    <drugCode2></drugCode2>  
                    <useLevel>2</useLevel>  
                    <drugName>水飞蓟宾</drugName>
                  </row>
                 </DrugInfoList>
                </function>
                ";
  XElement xe = XElement.Parse(xml);val element=from el in xe.Elements("DrugInfoList").Elements()
           select el;
element.ToList<XElement>().ForEach(x =>
  {Console.WriteLine("NodeName="+x.Name);
    if(x.HasElements)
    {
      x.ToList<XElement>().ForEach(xc =>
       {
      Console.WriteLine("Name:"+xc.Name+",Value:"+xc.Value);
       }
     }}

Answer 5

你，你还up..up也不是这么用的啊。

Answer 6

ls的都对，再补充个反序列化的。[XmlRoot(ElementName = "function")]
public class FunctionInfo
{
    public string function_id { get; set; }
    public string out_code { get; set; }
    public string out_text { get; set; }
    public string rows { get; set; }
    [XmlArray(ElementName = "DrugInfoList")]
    [XmlArrayItem(ElementName = "row", Type=typeof(DrugInfo))]
    public List<DrugInfo> drugInfoList { get; set; }
}public class DrugInfo
{
    public string classCode { get; set; }
    public string drugType { get; set; }
    public string drugCode { get; set; }
    public string drugCode2 { get; set; }
    public string useLevel { get; set; }
    public string drugName { get; set; }
}--------------------------------------------
FunctionInfo funcInfo = null;
using (var fs = File.OpenRead("XmlFile1.xml"))
{
    var xmlSer = new XmlSerializer(typeof(FunctionInfo));
    funcInfo = (FunctionInfo)xmlSer.Deserialize(fs);
}

Answer 7

如果仅仅是读取的话应该使用xmlreader对象，效率比较高，使用文件流的方式只读操作。
如果需要修改或新建，建议使用xmldocment对象，比较方便操作。

Answer 8

为什么改成下面的代码会报错？
string xml = @"
                <EP>
                    <q>
                    <qc>1</qc>
                    <no>1</no>
                    <fno>part 1</fno>
                    <sno>section A</sno>
                    <qb>Are you ready? A ok  B not ok</qb>
                    <qa></qa>
                    </q>
                    <q>
                    <qc>1</qc>
                    <no>1</no>
                    <fno>part 2</fno>
                    <sno>section B</sno>
                    <qb>Are you ready? A ok  B not ok</qb>
                    <qa></qa>
                    </q>
                    <q>
                    <qc>1</qc>
                    <no>1</no>
                    <fno>part 3</fno>
                    <sno>section C</sno>
                    <qb>Are you ready? A ok  B not ok</qb>
                    <qa></qa>
                    </q>
                </EP>
                ";
            XElement xe = XElement.Parse(xml);
            var fno = xe.Element("fno").Value;
            var sno = xe.Element("sno").Value;
            var qbs = xe.Descendants("qb").Select(x => x.Value);
            MessageBox.Show(fno);
            MessageBox.Show(sno);
            foreach (var i in qbs)
            {
                MessageBox.Show(i);
            }