我使用了以下类:
namespace defaultNamespace
{
...
public class DataModel
{
}
public class Report01
{get; set;}
public class Report02
{get; set;}
}
下面我有一个创建XML的方法。
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,它们有一个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);
当我尝试创建这样的XML时:
ObjectToXml(objs, "c:\12\objs.xml");
ObjectToXml(objs2, "c:\12\objs2.xml");
我看到这个例外:
不应使用类型"Report01(或Report02)"。使用XmlInclude或SoapInclude属性来指定未知的类型静态。
我该如何解决这个问题?
这是因为您的response.GetType()
实际上返回了List<object>
类型,然后您尝试序列化不期望的类型。Object
对您的类型一无所知,Object
的序列化程序无法序列化您的未知类型。
您可以对报告使用BaseClass,并使用XmlInclude
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");