Question

我想将一个XML文件用于C＃Objects。我的对象如下

[Serializable]
[XmlRoot(ElementName = "Collection")]
public class Collection
{
    public Collection() 
    {
        Artiesten = new List<Artiest>();
        Albums = new List<Album>();
        Nummers = new List<Nummer>();
    }
    [XmlElement("Artiesten")]
    public List<Artiest> Artiesten { get; set; }
    [XmlElement("Albums")]
    public List<Album> Albums { get; set; }
    [XmlElement("Nummers")]
    public List<Nummer> Nummers { get; set; }

}

[Serializable]
public class Artiest
{
    [XmlAttribute("artiestid")]
    public int ArtiestId { get; set; }
    [XmlElement(ElementName = "Naam")]
    public String Naam { get; set; }
    [XmlElement(ElementName = "Albums")]
    public List<Album> Albums { get; set; } 
}

[Serializable]
public class Nummer
{
[XmlAttribute("nummerid")]
    public int NummerId { get; set; }
    [XmlElement(ElementName = "titel")]
    public String Titel { get; set; }
    [XmlElement(ElementName = "duur")]
    public String Duration { get; set; }
}

我的XML是这样的：

<Collection xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"     xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Artiesten>
    <Artiest artiestid="1">
      <Naam>Harry</Naam>
      <Albums>
        <Album albumid="1">
          <Titel>Album1</Titel>
          <prijs valuta="Euro">19.99</prijs>
          <uitgiftejaar>1999</uitgiftejaar>
          <Nummers>
            <Nummer nummerid="1">
              <titel>happy Sundays</titel>
              <duur>PT02M02S</duur>
            </Nummer>
          </Nummers>
        </Album>
      </Albums>
    </Artiest>
  </Artiesten>
  <Albums>
    <Album albumid="1">
      <Titel>Album1</Titel>
      <prijs valuta="Euro">19.99</prijs>
      <uitgiftejaar>1999</uitgiftejaar>
      <Nummers>
        <Nummer nummerid="1">
          <titel>Happy Sundays</titel>
          <duur>PT02M02S</duur>
        </Nummer>
      </Nummers>
    </Album>
  </Albums>
  <Nummers>
    <Nummer nummerid="1">
      <titel>Happy Sundays</titel>
      <duur>PT02M02S</duur>
    </Nummer>
  </Nummers>
</Collection>

我正试图这样做：

XDocument doc = XDocument.Load(file);
                XmlSerializer xmlSerializer = new XmlSerializer(typeof(Collection));
                using (var reader = doc.Root.CreateReader())
                {
                    Collection collection = (Collection) xmlSerializer.Deserialize(reader);

                }

由于某种原因，我找不到Collection对象中的列表都是空的。调试显示XDocument中加载的文件有效。

编辑：我设法缩小了问题的范围。它确实正确地反序列化列表，只有那些列表中的对象的所有属性都是空的。

Answer 1

找到我的答案

我必须编辑我的列表属性：

    [XmlElement("Artiesten", typeof(List<Artiest>))]
    public List<Artiest> Artiesten { get; set; }
    [XmlElement("Albums", typeof(List<Album>))]
    public List<Album> Albums { get; set; }
    [XmlElement("Nummers", typeof(List<Nummer>))]
    public List<Nummer> Nummers { get; set; }

Answer 2

Collection collection = null;
string path = "file.xml";

XmlSerializer serializer = new XmlSerializer(typeof(Collection));

StreamReader reader = new StreamReader(path);
collection = (Collection)serializer.Deserialize(reader);
reader.Close();

Answer 3

您需要重新移动列表中的[XmlElement]标记。否则，它使用不同的XML结构。

例如，不是将所有Artiest个对象嵌套在一个Artiesten元素（这是您当前的XML）中，而是将它们实际设置为彼此相邻：

  <Artiesten artiestid="1">
      <Naam>Harry</Naam>
      <Albums>
        <Album albumid="1">
          <Titel>Album1</Titel>
          <prijs valuta="Euro">19.99</prijs>
          <uitgiftejaar>1999</uitgiftejaar>
          <Nummers>
            <Nummer nummerid="1">
              <titel>happy Sundays</titel>
              <duur>PT02M02S</duur>
            </Nummer>
          </Nummers>
        </Album>
      </Albums>
  </Artiesten>
  <Artiesten artiestid="2">
      <Naam>Harry</Naam>
      <Albums>
        <Album albumid="1">
          <Titel>Album1</Titel>
          <prijs valuta="Euro">19.99</prijs>
          <uitgiftejaar>1999</uitgiftejaar>
          <Nummers>
            <Nummer nummerid="1">
              <titel>happy Sundays</titel>
              <duur>PT02M02S</duur>
            </Nummer>
          </Nummers>
        </Album>
      </Albums>
  </Artiesten>
  <Artiesten artiestid="3">
      <Naam>Harry</Naam>
      <Albums>
        <Album albumid="1">
          <Titel>Album1</Titel>
          <prijs valuta="Euro">19.99</prijs>
          <uitgiftejaar>1999</uitgiftejaar>
          <Nummers>
            <Nummer nummerid="1">
              <titel>happy Sundays</titel>
              <duur>PT02M02S</duur>
            </Nummer>
          </Nummers>
        </Album>
      </Albums>
  </Artiesten>

因此，请尝试重新定义您的课程：

[Serializable]
[XmlRoot(ElementName = "Collection")]
public class Collection
{
    public Collection() 
    {
        Artiesten = new List<Artiest>();
        Albums = new List<Album>();
        Nummers = new List<Nummer>();
    }

    public List<Artiest> Artiesten { get; set; }
    public List<Album> Albums { get; set; }
    public List<Nummer> Nummers { get; set; }

}

[Serializable]
public class Artiest
{
    [XmlAttribute("artiestid")]
    public int ArtiestId { get; set; }
    [XmlElement(ElementName = "Naam")]
    public String Naam { get; set; }

    public List<Album> Albums { get; set; } 
}

Answer 4

你想要XmlArray，而不是XmlElement：

[XmlArray("Artiesten")]
[XmlArrayItem("Artiest")]
public List<Artiest> ...

实际上这是列表的默认行为，因此您也只需完全删除该属性。

反序列化返回空对象

4 个答案: