给出以下XML:
<RESPONSE version="10">
<RESULTS MyAttribute="1" />
</RESPONSE>
我可以像这样反序列化:
[XmlRoot("RESPONSE")]
public class Response
{
[XmlElement(ElementName = "RESULTS")]
public Results Results {get;set;}
}
public class Results
{
[XmlAttribute(AttributeName = "MyAttribute")]
public bool MyAttribute { get; set; }
}
var serializer = new XmlSerializer(typeof(Response));
var result = (Response)serializer.Deserialize(new StringReader(xml));
这很有效。 但问题是我想要反序列化以下XML,并且我不想复制Response类,我想使用通用方法。
<RESPONSE version="10">
<SOCCER AnotherAttribute="1" />
</RESPONSE>
我尝试过以下操作:
[XmlRoot("Response")]
public class Response<T>
{
public T Data {get;set;}
}
public class SoccerRoot
{
[XmlElement(ElementName = "SOCCER")]
public class Soccer {get;set;}
}
public class Soccer
{
[XmlAttribute(AttributeName = "AnotherAttribute")]
public bool AnotherAttribute {get;set;}
}
var serializer = new XmlSerializer(typeof(Response<SoccerRoot>));
var result = (Response<SoccerRoot>)serializer.Deserialize(new StringReader(xml));
但这不起作用,我也试图继承Response类但没有运气。
我收到以下错误:<xmlns='' > was not expected.
我在这里是正确的方式,还是在反序列化XML文件时不可能使用通用方法?
答案 0 :(得分:3)
使用Response<SoccerRoot>
,预期的xml布局为:
<RESPONSE>
<Data>
<SOCCER AnotherAttribute=...>
与匹配。无法使用属性自定义T Data
的元素名称。
选项:
有多个根对象(SOCCER
,RESULTS
等各一个):
[XmlRoot("RESPONSE")]
public class SoccerRoot {
[XmlElement("SOCCER")]
public Soccer Soccer {get;set;}
}
[XmlRoot("RESPONSE")]
public class ResultsRoot {
[XmlElement("RESULTS")]
public Results Results {get;set;}
}
有一个具有多个第一级子项的根对象(一个用于SOCCER
,一个用于RESULTS
等等 - 每个类型
[XmlRoot("RESPONSE")]
public class Response {
[XmlElement("RESULTS")]
public Results Results {get;set;}
[XmlElement("SOCCER")]
public Soccer Soccer {get;set;}
}
使用泛型方法,但丢失了中间对象,因此您Response<Soccer>
(完全删除SoccerRoot
),并使用XmlAttributeOverrides
自定义元素名称确保缓存XmlSerializer
生成的(否则会以惊人的速度泄漏)
坦率地说,我会选择前两个选项中的任何一个,而单独留下第三个选项。
答案 1 :(得分:1)
尝试创建一个包含对象RESULTS和SOCCER的公共类,然后根据数据,一个将为null,其他将包含值。
class Program
{
static void Main(string[] args)
{
//
var xml = "<RESPONSE version=\"1\" ><SOCCER AnotherAttribute =\"1\" /></RESPONSE>";
var serializer = new XmlSerializer(typeof(Response<int>));
var result = (Response<int>)serializer.Deserialize(new StringReader(xml));
xml = "<RESPONSE version=\"10\"><RESULTS MyAttribute = \"1\" /></RESPONSE >";
result = (Response<int>)serializer.Deserialize(new StringReader(xml));
Console.ReadLine();
}
}
[XmlRoot("RESPONSE")]
public class Response<T>
{
[XmlAttribute(AttributeName = "version")]
public T Version { get; set; }
[XmlElement(ElementName = "SOCCER")]
public SoccerRoot SoccerRoot { get; set; }
[XmlElement(ElementName = "RESULTS")]
public Results Results { get; set; }
}
public class SoccerRoot
{
[XmlAttribute(AttributeName = "AnotherAttribute")]
public int AnotherAttribute { get; set; }
}
public class Results
{
[XmlAttribute(AttributeName = "MyAttribute")]
public bool MyAttribute { get; set; }
}