C# 无法反序列化此XML
只是为了好玩,我在玩Last.fm的API。他们为顶级艺术家返回的XML文件的结构如下:C# 无法反序列化此XML,c#,xml,xml-deserialization,last.fm,C#,Xml,Xml Deserialization,Last.fm,只是为了好玩,我在玩Last.fm的API。他们为顶级艺术家返回的XML文件的结构如下: <lfm status="ok"> <topartists user="xbonez" type="overall"> <artist rank="1"> <name>Evanescence</name> <playcount>4618</playcount> <mbid>f4a31f0a-
<lfm status="ok">
<topartists user="xbonez" type="overall">
<artist rank="1">
<name>Evanescence</name>
<playcount>4618</playcount>
<mbid>f4a31f0a-51dd-4fa7-986d-3095c40c5ed9</mbid>
<url>http://www.last.fm/music/Evanescence</url>
<streamable>1</streamable>
<image size="small">http://userserve-ak.last.fm/serve/34/48488613.png</image>
<image size="medium">http://userserve-ak.last.fm/serve/64/48488613.png</image>
<image size="large">http://userserve-ak.last.fm/serve/126/48488613.png</image>
<image size="extralarge">http://userserve-ak.last.fm/serve/252/48488613.png</image>
<image size="mega">http://userserve-ak.last.fm/serve/500/48488613/Evanescence++PNG.png</image>
</artist>
</topartists>
</lfm>
然后我使用以下代码反序列化:
string XmlFile = "artists.xml";
XmlSerializer serializer = new XmlSerializer(typeof(lfmStatus));
lfmStatus LoadFile;
using (Stream reader = new FileStream(XmlFile, FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
{
try
{
Console.WriteLine("Beginning deserialization.");
// Call the Deserialize method to restore the object's state.
LoadFile = (lfmStatus)serializer.Deserialize(reader);
return LoadFile.TopArtists;
}
现在,如果XML没有包含所有艺术家的topartists标记,那么这段代码对XML非常有用。但既然这样,我该如何修改代码来处理它呢?我假设我需要添加另一个类。您缺少一些类型的属性 有关更多详细信息,请参阅 您还缺少
topartists
元素的类型
如果我是你,我会得到XML模式,然后使用xsd.exe生成C#类,并从中进行修改。如果您确实找不到,它还可以基于XML推断模式,这将根据输入XML为您提供一个可解析的结果。要查看您编写的反序列化响应XML的正确代码,可以使用XSD。打开VS命令提示符并给出XSD LastFM.xml,该文件将生成和XSD文件。现在给出XSD LastFM.XSD,它将生成一个CS文件。把这个和你写的比较一下,检查一下你是否犯了错误
[Serializable()]
public class Artists
{
[XmlElement("name")]
public string Name { get; set; }
[XmlElement("playcount")]
public int playcount { get; set; }
[XmlElement("url")]
public string url { get; set; }
[XmlElement("streamable")]
public int streamable { get; set; }
[XmlElement("image")]
public string image { get; set; }
}
string XmlFile = "artists.xml";
XmlSerializer serializer = new XmlSerializer(typeof(lfmStatus));
lfmStatus LoadFile;
using (Stream reader = new FileStream(XmlFile, FileMode.Open, FileAccess.Read, FileShare.ReadWrite))
{
try
{
Console.WriteLine("Beginning deserialization.");
// Call the Deserialize method to restore the object's state.
LoadFile = (lfmStatus)serializer.Deserialize(reader);
return LoadFile.TopArtists;
}