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c#xml操作

news 来源：原创 2024/5/2 13:54:24

简单的xml操作--解析技能xml

xml文件

<skills>
  
    <skill>
    <id>1</id>
    <name lang="cn">大荒囚天指</name>
    <demage>100</demage>     
  </skill>
  
  <skill>
    <id>2</id>
    <name lang="en">绝对零度</name>
    <demage>200</demage>
  </skill>
    
  <skill>
      <id>3</id>
      <name lang="ger">青龙魂</name>
      <demage>1000</demage>
  </skill>

</skills>

创建对应的skill类，重写了ToString方法方便输出

class Skill
    {
        public int ID { get; set; }
        public string Name { get; set; }
        public string Lang { get; set; }
        public int Demage { get; set; }

        public override string ToString()
        {
            return string.Format("Id:{0},Name:{1},Lang:{2},Demage:{3}", ID, Name, Lang, Demage);
        }
    }

 1 using System;
 2 using System.Collections.Generic;
 3 using System.Linq;
 4 using System.Text;
 5 using System.Threading.Tasks;
 6 using System.Xml;
 7 
 8 namespace xml操作
 9 {
10     class Program
11     {
12         static void Main(string[] args)
13         {
14             List<Skill> skillList = new List<Skill>();
15 
16             XmlDocument xmlDoc = new XmlDocument();
17             xmlDoc.Load("skillInfo.txt");
18 
19             //根节点
20             XmlNode root = xmlDoc.FirstChild;
21 
22             XmlNodeList skillsNodeList = root.ChildNodes;
23             foreach (XmlNode skillNode in skillsNodeList)
24             {
25                 Skill skill = new Skill();
26                 foreach (XmlNode filedNode in skillNode.ChildNodes)
27                 {
28                     if (filedNode.Name == "id")
29                     {
30                         skill.ID = int.Parse( filedNode.InnerText);
31                     }
32                     else if (filedNode.Name == "name")
33                     {
34                         skill.Name = filedNode.InnerText;
35                         skill.Lang = filedNode.Attributes[0].Value;
36                     }
37                     else if (filedNode.Name == "demage")
38                     {
39                         skill.Demage = int.Parse(filedNode.InnerText);
40                     }
41                 }
42                 skillList.Add(skill);
43             }
44 
45             foreach (var item in skillList)
46             {
47                 Console.WriteLine(item);
48             }
49         }
50     }
51 }