C# 如何使用XmlInclude动态指定类型?
我使用的课程有:C# 如何使用XmlInclude动态指定类型?,c#,xml,object,xmlinclude,C#,Xml,Object,Xmlinclude,我使用的课程有: namespace defaultNamespace { ... public class DataModel { } public class Report01 {get; set;} public class Report02 {get; set;} } 我有一个方法可以创建下面的XML public XmlDocument ObjectToXml(object response, strin
namespace defaultNamespace
{
...
public class DataModel
{
}
public class Report01
{get; set;}
public class Report02
{get; set;}
}
我有一个方法可以创建下面的XML
public XmlDocument ObjectToXml(object response, string OutputPath)
{
Type type = response.GetType();
System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(type);
MemoryStream stream = new MemoryStream();
StreamWriter writer = new StreamWriter(stream, Encoding.UTF8);
serializer.Serialize(writer, response);
XmlDocument xmldoc = new XmlDocument();
stream.Position = 0;
StreamReader sReader = new StreamReader(stream);
xmldoc.Load(sReader);
stream.Position = 0;
string tmpPath = OutputPath;
while (File.Exists(tmpPath))
{
File.Delete(tmpPath);
}
xmldoc.Save(tmpPath);
return xmldoc;
}
我有两个列表,分别有Report01和Report02对象
List<object> objs = new List<object>();
List<object> objs2 = new List<object>();
Report01 obj = new Report01();
obj.prop1 = "aa";
obj.prop2 = "bb";
objs.Add(obj);
Report02 obj2 = new Report02();
obj2.prop1 = "cc";
obj2.prop2 = "dd";
objs2.Add(obj2);
我看到这个例外:
[XmlInclude(typeof(Report01)]
[XmlInclude(typeof(Report02)]
public class BaseClass { }
public class Report01 : BaseClass { ... }
public class Report02 : BaseClass { ... }
List<BaseClass> objs = new List<BaseClass>();
List<BaseClass> objs2 = new List<BaseClass>();
// fill collections here
ObjectToXml(objs, "c:\\12\\objs.xml");
ObjectToXml(objs2, "c:\\12\\objs2.xml");
不应使用类型“Report01(或Report02)”。使用xmlclude
或SoapInclude属性指定未知的类型
静态地
如何解决此问题?这是因为您的
响应。GetType()
实际上返回列表
类型,然后您尝试序列化非预期类型<代码>对象不了解您的类型,对象的序列化程序
无法序列化未知类型
您可以将基类用于报告和xmlclude
来解决此异常:
[XmlInclude(typeof(Report01)]
[XmlInclude(typeof(Report02)]
public class BaseClass { }
public class Report01 : BaseClass { ... }
public class Report02 : BaseClass { ... }
List<BaseClass> objs = new List<BaseClass>();
List<BaseClass> objs2 = new List<BaseClass>();
// fill collections here
ObjectToXml(objs, "c:\\12\\objs.xml");
ObjectToXml(objs2, "c:\\12\\objs2.xml");
[xmlclude(typeof(Report01)]
[xmlclude(typeof(Report02)]
公共类基类{}
公共类Report01:基类{…}
公共类报表02:基类{…}
List objs=new List();
List objs2=新列表();
//在这里填写集合
ObjectToXml(objs,“c:\\12\\objs.xml”);
ObjectToXml(objs2,“c:\\12\\objs2.xml”);
所以,您说我需要创建一个Report01和Report02将继承的基类。对吗?是的,因为您不能向对象添加xmlcludeAttribute
。或者您可以声明List
和List
,而不是List
。