获取XML元素和子元素到列表中

Question

鉴于以下XML，如何使用以下实现将每条记录读入对象列表（记录列表）？

class Record 
{ 
    public string Name {get; set;} 
    public List<Skill> Skills {get; set;} 
} 

class Skill 
{ 
    public string SkillName {get; set;} 
    public string SkillType {get; set;} 
} 

XML

<records> 
    <record> 
     <name>John</name> 
     <skills> 
     <skill> 
      <skillname>Tennis</skillname> 
      <skilltype>Sport</skilltype> 
     </skill> 
     <skill> 
      <skillname>Violin</skillname> 
      <skilltype>Music</skilltype> 
     </skill> 
     </skills> 
    </record> 
    <record> 
     <name>Juanita</name> 
     <skills> 
     <skill> 
      <skillname>Football</skillname> 
      <skilltype>Sport</skilltype> 
     </skill> 
     <skill> 
      <skillname>Pottery</skillname> 
      <skilltype>Craft</skilltype> 
     </skill> 
     </skills> 
    </record> 
</records> 

我试图让子元素（技能）都没有成功。

List<Records> records; 
DataSet dstRecords = new DataSet(); 
dstRecords.ReadXml(file); 

foreach (DataRow row in dstRecords.Tables[0].Rows) 
      { 

       Record record = new Record 
       { 
        Name = Convert.ToString(row["name"]), 
        Skills = skills 
       }; 
       records.Add(records); 
      } 

Answer 1

XDocument xDoc = XDocument.Parse(xmlstring); //or XDocument.Load(filename); 
var list = xDoc.Descendants("record") 
      .Select(r => new Record 
      { 
       Name = (string)r.Element("name"), 
       Skills = r.Descendants("skill") 
          .Select(s=>new Skill{ 
           SkillName = (string)s.Element("skillname"), 
           SkillType = (string)s.Element("skilltype"), 
          }) 
          .ToList() 
      }) 
      .ToList(); 

Answer 2

假设你想LINQ到XML，您只需将 “链”，以Element和Elements（或Descendants）所需的呼叫：

// assuming XDocument xdoc = ... 
var records = from r in xdoc.Root.Elements("record") 
       select new Record 
       { 
        Name = (string)r.Element("name"), 
        Skills = (from s in r.Element("skills").Elements("skill") 
          select new Skill 
          { 
           SkillName = (string)s.Element("skillname"), 
           SkillType = (string)s.Element("skilltype") 
          }).ToList() 
       };