我有我使用的这些类:
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;
}
我有两个具有 Report01 和 Report02 对象的列表。
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);
当我尝试像这样创建 XML 时:
ObjectToXml(objs, "c:\\12\\objs.xml");
ObjectToXml(objs2, "c:\\12\\objs2.xml");
我看到了这个例外:
不需要类型“Report01(或 Report02)”。使用 XmlInclude 或 SoapInclude 属性指定静态未知的类型。
我怎么解决这个问题?