什麼會導致此自定義XML ModelBinder不反序列化我的XML POST？

Question

模型

public class SimpleUser 
{ 
    public string FirstName { get; set; } 
    public string LastName { get; set; } 
    public string UserName { get; set; } 
    public int Role { get; set; } 
    public bool isActive { get; set; } 

    public string Groups { get; set; } 
} 

的BinderProvider

public class SimpleUserProvider : IModelBinderProvider 
{ 
    public IModelBinder GetBinder(Type modelType) 
    { 
     var contentType = HttpContext.Current.Request.ContentType; 

     if (string.Compare(contentType, @"text/xml", StringComparison.OrdinalIgnoreCase) != 0) 
     { 
      return null; 
     } 

     return new SimpleUserBinder(); 
    } 
} 

的ModelBinder的

public class SimpleUserBinder : IModelBinder 
{ 
    public object BindModel(ControllerContext controllerContext, ModelBindingContext bindingContext) 
    { 
     var modelType = bindingContext.ModelType; 
     var serializer = new XmlSerializer(modelType); 

     var inputStream = controllerContext.HttpContext.Request.InputStream; 

     return serializer.Deserialize(inputStream); 
    } 
} 

的的Application_Start（）中Global.a sax.cs

protected void Application_Start() 
    { 
     AreaRegistration.RegisterAllAreas(); 

     ModelBinderProviders.BinderProviders.Add(new SimpleUserProvider()); 

     RegisterGlobalFilters(GlobalFilters.Filters); 
     RegisterRoutes(RouteTable.Routes); 
    } 

行動

[HttpPost] 
    public ActionResult Create(SimpleUser u) 
    { 
     //simple output for testing bind 

     return Content(u.FirstName + ", " + u.LastName + ", " + u.UserName + ", " + u.Role.ToString() + ", " + u.isActive + ", {" + u.Groups + "}", "text/plain"); 
    } 

然而，當我張貼含有這種XML格式的 「文本/ xml」 的要求：

<SimpleUser> 
<FirstName>John</FirstName> 
<LastName>Smith</LastName> 
<UserName>jsmith@someland.com</UserName> 
<Role>3</Role> 
<isActive>true</isActive> 
</SimpleUser> 

我得到的返回是：

，0，假，{}

我跟着this post，我缺少的是在這裏嗎？

Answer 1

我在這裏錯過了什麼？

你缺少消費之前將其復位流：

var inputStream = controllerContext.HttpContext.Request.InputStream; 
inputStream.Position = 0; 
return serializer.Deserialize(inputStream); 

或使用XmlReader：

using (var inputStream = controllerContext.HttpContext.Request.InputStream) 
using (var reader = XmlReader.Create(inputStream)) 
{ 
    return serializer.Deserialize(reader); 
}