标签:
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 }
标签:
原文地址:http://www.cnblogs.com/zhangbaochong/p/4924558.html