对XMl文件的操作

一.在工程目录下，新建txt文件取名Skills，修改其属性运行时复制到工作目录中；

右键选择Open with

选择Xml方便编辑txt内容

二.新建一个SkillInfo类，用来存储Skill信息

class SkillInfo
{
    public int ID { get; set; }
    public string Name { get; set; }
    public int damage { set; get; }
    public string Language { get; set; }
    public override string ToString()
    {
        return string.Format("ID:" + ID + ",Name:" + Name + ",damage:" + damage + ",Language:" + Language);
    }
}

static void Main(string[] args)
        {
            List<SkillInfo> SkillList = new List<SkillInfo>();
            XmlDocument xmlDoc = new XmlDocument();
            xmlDoc.Load("Skills.txt");
            XmlNode skills = xmlDoc.FirstChild;
            XmlNodeList skillList = skills.ChildNodes;
            foreach (XmlNode skill in skillList)
            {
                XmlNodeList skillInfo = skill.ChildNodes;
                SkillInfo skillObject = new SkillInfo();
                foreach (XmlNode info in skillInfo)
                {
                    if (info.Name== "ID")
                    {
                        skillObject.ID =int.Parse(info.InnerText);
                    }
                    if (info.Name == "Name")
                    {
                        skillObject.Name = info.InnerText;
                        skillObject.Language = info.Attributes[0].Value;
                    }
                    if (info.Name == "damage")
                    {
                        skillObject.damage = int.Parse(info.InnerText);
                    }
                }
                SkillList.Add(skillObject);
            }
            foreach(SkillInfo list in SkillList)
            {
                Console.WriteLine(list.ToString());
            }
            Console.ReadKey();
        }

Skills.txt文本编写格式如下：

结果如下：