在复杂对象中反序列化xml

时间:2012-03-26 11:00:42

标签: c# asp.net-mvc-3 xml-parsing xml-deserialization

我无法理解为什么对象为空:

        WebClient browse = new WebClient();
        StreamReader res = new StreamReader(browse.OpenRead("http://ws.audioscrobbler.com/2.0/?method=track.getinfo&api_key=b25b959554ed76058ac220b7b2e0a026&artist=cher&track=believe"));
        string result = res.ReadToEnd();

        XmlDocument xmltrackinfo = new XmlDocument();
        xmltrackinfo.InnerXml = result;



        XmlRootAttribute xRoot = new XmlRootAttribute();
        xRoot.ElementName = "lfm";
        xRoot.IsNullable = true;

        XmlSerializer xs = new XmlSerializer(typeof(fm), xRoot);

        fm rez = (fm) xs.Deserialize(new MemoryStream(Encoding.UTF8.GetBytes(result)));

对象模型:

[Serializable()]
[XmlRoot(ElementName = "lfm", IsNullable = true)]
public class fm
{
    [XmlElement("lfm")]
    public Track lfm { get; set; }
}

[Serializable()]
[XmlRoot(ElementName = "artist", IsNullable = true)]
public class artist
{
    public string name { get; set; }
    public string mbid { get; set; }
    public string url { get; set; }
}

[Serializable()]
[XmlRoot(ElementName = "album", IsNullable = true)]
public class album
{
    public string artist { get; set; }
    public string title { get; set; }
    public string mbid { get; set; }
    public string url { get; set; }
    public List<string> image { get; set; }
}

[Serializable()]
[XmlRoot(ElementName = "tag", IsNullable = true)]
public class tag
{
    public string name { get; set; }
    public string url { get; set; }

}

[Serializable()]
[XmlRoot(ElementName = "wiki", IsNullable = true)]
public class wiki
{
    public string summary { get; set; }
    public string content { get; set; }
}

[Serializable()]
[XmlRoot(ElementName = "track", IsNullable = true)]
public class Track
{
    public string id { get; set; }
    public string name { get; set; }
    public string mbid { get; set; }
    public string url { get; set; }
    public string duration { get; set; }
    public string streamable { get; set; }
    public string listeners { get; set; }
    public string playcount { get; set; }
    public artist artist { get; set; }
    public album album { get; set; }
    public List<tag> toptags { get; set; }
    public wiki wiki { get; set; }

}

和XML:

http://ws.audioscrobbler.com/2.0/?method=track.getinfo&api_key=b25b959554ed76058ac220b7b2e0a026&artist=cher&track=believe

那我该怎么办?

2 个答案:

答案 0 :(得分:2)

尝试将fm课程重命名为lfm

public class lfm
{
    public Track track { get; set; }
}

然后你也可以摆脱xRoot变量:

XmlSerializer xs = new XmlSerializer(typeof(lfm));
lfm rez = (lfm) xs.Deserialize(new MemoryStream(Encoding.UTF8.GetBytes(result)));

此外,您不需要[Serializable]属性。这用于二进制序列化,XmlSerializer类完全忽略它。

答案 1 :(得分:1)

lfm类的fm属性必须跟踪作为其XmlElement:

[Serializable()]
[XmlRoot(ElementName = "lfm", IsNullable = true)]
public class fm
{
    [XmlElement("track")]
    public Track lfm { get; set; }
}