异常的XmlSerializer的一个动态ExpandoObject

我有一个类，并在课堂上有，当我打电话序列化的同一类然后把exception-异常的XmlSerializer的一个动态ExpandoObject

要XML序列化，其继承的类型属性类型动态列表从IEnumerable必须有继承层次结构的所有级别的Add（System.Object）实现。 System.Dynamic.ExpandoObject不执行添加（System.Object）。

Class架构 -

public class TestClass 
{ 
    public string Property1{ get; set; } 
    public string Property2{ get; set; } 
    public string Property3{ get; set; } 
    public string Property4{ get; set; } 

    public List<dynamic> ProductList { get; set; } 
}

XmlSerializer的XmlSerializer的新= XmlSerializer的（TestClass.GetType（））;

 using (StringWriter textWriter = new StringWriter()) 
     { 
      xmlSerializer.Serialize(textWriter, Obj); 
      string xmlString=textWriter.ToString(); 
     }

来源

2017-05-25 amethianil

-1

请使用一个模型类，而不是动态的序列化，因为类型需要动态进行定义。

我所做的每一个属性分配正确的价值观，改变了TestClass.GetType（）来obj.GetType（）和打印的TextWriter。请检查下面的代码：

的ModelClass：

public class ModelClass 
{ 
    public string Name { get; set; } 
    public string Search { get; set; } 
}

的识别TestClass：

public class TestClass 
{ 
    public string Property1 { get; set; } 
    public string Property2 { get; set; } 
    public string Property3 { get; set; } 
    public string Property4 { get; set; } 

    public List<ModelClass> ProductList { get; set; } 
}

的类执行程序：

class Program 
{ 
    public static void Main(string[] args) 
    { 
     List<ModelClass> productList = new List<ModelClass>(); 

     var product1 = new ModelClass(); 
     product1.Name = "Name1"; 
     product1.Search = "Search1"; 

     productList.Add(product1); 

     var product2 = new ModelClass(); 
     product2.Name = "Name2"; 
     product2.Search = "Search2"; 

     productList.Add(product2); 

     var product3 = new ModelClass(); 
     product3.Name = "Name3"; 
     product3.Search = "Search3"; 

     productList.Add(product3); 

     TestClass obj = new TestClass(); 

     obj.Property1 = "p1"; 
     obj.Property2 = "p2"; 
     obj.Property3 = "p3"; 
     obj.Property4 = "p4"; 
     obj.ProductList = productList; 

     XmlSerializer xmlSerializer = new XmlSerializer(obj.GetType()); 

     using (StringWriter textWriter = new StringWriter()) 
     { 
      xmlSerializer.Serialize(textWriter, obj); 
      string xmlString = textWriter.ToString(); 
      Console.WriteLine(textWriter); 
     } 

     Console.ReadKey(); 
    } 
}

来源

2017-05-25 09:21:23 ViVi

好的谢谢。但我的情况产品列表的属性不是一个字符串列表。请检查我的回答 – amethianil

@amethianil你仍然面临例外吗？ – ViVi

List productList = new List（）;

 dynamic product1 = new ExpandoObject(); 
     product1.Name = "Name1"; 
     product1.Search = "Search1"; 

     productList.Add(product1); 

     dynamic product2 = new ExpandoObject(); 
     product2.Name = "Name2"; 
     product2.Search = "Search2"; 

     productList.Add(product2); 

     dynamic product3 = new ExpandoObject(); 
     product3.Name = "Name3"; 
     product3.Search = "Search3"; 

     productList.Add(product3); 

     TestClass obj = new TestClass(); 

     obj.Property1 = "p1"; 
     obj.Property2 = "p2"; 
     obj.Property3 = "p3"; 
     obj.Property4 = "p4"; 
     obj.ProductList = productList; 

     XmlSerializer xmlSerializeraa = new XmlSerializer(obj.GetType()); 

     using (StringWriter textWriter = new StringWriter()) 
     { 
      xmlSerializeraa.Serialize(textWriter, obj); 
      string xmlStringOut = textWriter.ToString(); 
      Console.WriteLine(textWriter); 
     }

来源

2017-05-25 09:45:49 amethianil

异常的XmlSerializer的一个动态ExpandoObject

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