在 C# 中解析 XML 数据并显示到 ListBox

Question

我正在尝试使用 Visual Studio 解析 C# 中的 XML 文件并在 ListBox 中显示数据，但在处理嵌套的 XML 文件时我不知道如何解析它。

这是来自 XML 文件的代码：

<?xml version="1.0" encoding="utf-8" ?>
<!DOCTYPE root [
  <!ELEMENT root (Persons*)>
  <!ELEMENT Persons (name)>
  <!ELEMENT IsMale (#PCDATA)>
  <!ELEMENT Age (#PCDATA)>
  <!ELEMENT Name (#PCDATA)>
  <!ELEMENT LikedPerson (name)>
 ]>
<root>
  <Persons name ="Bob">
    <IsMale>true</IsMale>
    <Age>30</Age>
    <LikedPerson name ="Iulia">
      <IsMale>false</IsMale>
      <Age>32</Age>
    </LikedPerson>
  </Persons>
</root>

我用 C# 编写的代码成功地只为我返回了每个人的姓名、性别和年龄，但我不知道如何编写以显示 person_liked：

private void LoadPersons()
    {
        XmlDocument doc = new XmlDocument();
        doc.Load("Baza_de_cunostinte.xml");

        foreach (XmlNode node in doc.DocumentElement) 
        {
            string name = node.Attributes[0].Value;
            int age = int.Parse(node["Age"].InnerText);
            bool isMale = bool.Parse(node["IsMale"].InnerText);

//          Persons likedPerson.name = Persons.node.Attributes[0].Value ?  
//          .....

            listBox.Items.Add(new Persons(name, age, isMale, likedPerson));
        }
    }

    private void listBox_SelectedIndexChanged(object sender, EventArgs e)
    {
        if (listBox.SelectedIndex != -1)
        {
            propertyGrid1.SelectedObject = listBox.SelectedItem;
        }
    }

这是 Persons.cs 的定义：

class Persons
{
    public string Name { get; private set; }
    public int Age { get; private set; }
    public bool IsMale { get; private set; }
    public Persons LikedPerson { get; private set; }

    public Persons(string name, int age, bool isMale, Persons likedPerson)
    {
        Name = name;
        Age = age;
        IsMale = isMale;
        LikedPerson = likedPerson;
    }
}

Answer 1

XmlSerializer mySerializer = new XmlSerializer(typeof(Persons));
// Create a FileStream or textreader to read the xml data.
FileStream myFileStream = new FileStream("xmldatafile.xml", FileMode.Open);

var person = (Persons)  mySerializer.Deserialize(myFileStream);

您还需要为 Persons 类添加不带参数的构造函数。

Answer 2

按照建议，最自然的方法是使用XmlSerializer，但要这样做，您必须稍微重构您的类：

[XmlType(Namespace="", TypeName="root")]
public class PersonCollection
{
    [XmlElement(Namespace="", ElementName="Persons")]
    public List<Persons> People { get; set; }
}

public class Persons
{
    [XmlAttribute(AttributeName="name")]
    public string Name { get; set; }
    public int Age { get; set; }
    public bool IsMale { get; set; }

    public Persons LikedPerson { get; set; }

    public Persons() { }

    public Persons(string name, int age, bool isMale, Persons likedPerson)
    {
        Name = name;
        Age = age;
        IsMale = isMale;
        LikedPerson = likedPerson;
    }
}

然后你可以这样做：

XmlSerializer ser = new XmlSerializer(typeof(PersonCollection));

PersonCollection pc = (PersonCollection)ser.Deserialize(File.OpenRead("Baza_de_cunostinte.xml"));
foreach (Persons p in pc.People)
{
   // you now have a fully populated object
}

pc.People 列表将包含您的 Persons 对象。

Answer 3

@user3063909,

1- 为 XML 定义使用 XSD。例如：

<xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
  <xs:element name="root">
    <xs:complexType>
      <xs:sequence>
        <xs:element name="Persons" maxOccurs="unbounded" minOccurs="0" type="Persons"/>
      </xs:sequence>
    </xs:complexType>
  </xs:element>
  <xs:complexType name="Persons">
    <xs:sequence>
      <xs:element type="xs:string" name="IsMale"/>
      <xs:element type="xs:int" name="Age"/>
      <xs:element name="LikedPerson" type="Persons"/>
    </xs:sequence>
    <xs:attribute type="xs:string" name="name" />
  </xs:complexType>
</xs:schema>   

2- Persons 类应如下所示：

namespace StackOverflow
{
    public class Root
    {
        [XmlElement("Persons")]
        public List<Persons> Persons { get; set; }
    }

    public class Persons
    {
        public string IsMale { get; set; }
        public int Age { get; set; }
        public Persons LikedPerson { get; set; }

        [XmlAttribute("Name")]
        public string Name { get; set; }
    }
}

3- 序列化器类：

namespace StackOverflow
{
    public class XmlSerializerHelper<T> where T : class 
    {
        private readonly XmlSerializer _serializer;

        public XmlSerializerHelper()
        {
            _serializer = new XmlSerializer(typeof(T));
        }

        public T BytesToObject(byte[] bytes)
        {
            using (var memoryStream = new MemoryStream(bytes))
            {
                using (var reader = new XmlTextReader(memoryStream))
                {
                    return (T)_serializer.Deserialize(reader);
                }
            }
        }
    }
}

4- 最后，这样称呼它：

var fileBytes = File.ReadAllBytes("C:/xml.xml");
var persons = new XmlSerializerHelper<Root>().BytesToObject(fileBytes);

结果将是带有人员列表的根类。

干杯。

Answer 4

您可以像现在一样获取 LikedPerson 节点并获取它的名称/年龄。为了避免代码重复，您可以创建一个方法，该方法接受 XmlNode，递归解析它并返回一个 Person。但更好的方法是使用XmlSerializer

foreach (XmlNode node in doc.DocumentElement) 
{
    string name = node.Attributes[0].Value;
    int age = int.Parse(node["Age"].InnerText);
    bool isMale = bool.Parse(node["IsMale"].InnerText);

    var likedPerson = node.SelectSingleNode("LikedPerson");

    if (likedPerson != null){
        string name = likedPerson.Attributes[0].Value;
        //age, gender, etc.        
    }        
}