Swagger/Swashbuckle无法识别请求模型中的JObject

Question

我的环境:Asp。. NET WebAPI， . NET Framework 4.5.2, Swashbuckle。核心5.6.0

由于某些原因，我的控制器必须继承之前相同的控制器，像这样

public class User100Controller : ApiController
{
[HttpGet]
[AllowAnonymous]
public virtual string Get()
{
return "1.0.0";
}
}
public class User101Controller : User100Controller
{
[HttpGet]
[AllowAnonymous]
public override string Get()
{
return "1.0.1";
}
}

直接运行，并确保ui页面显示正确

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但是如果我用参数添加Post Action, swagger ui不识别请求模型

UserModel和UserModelSex

/// <summary>
/// UserModel
/// </summary>
public class UserModel
{
/// <summary>
/// Name
/// </summary>
public string name { get; set; } = string.Empty;
/// <summary>
/// Age
/// </summary>
public int age { get; set; } = 0;
}
/// <summary>
/// UserModelSex
/// </summary>
public class UserModelSex : UserModel
{
/// <summary>
/// Sex
/// </summary>
public int sex { get; set; } = -1;
}

User100Controller

[HttpPost]
[SwaggerResponse(200, "success", typeof(UserModel))]
public virtual IHttpActionResult SaveUser([FromBody] UserModel model)
{
if (string.IsNullOrEmpty(model.name) || model.age == 0)
{
return Ok("error");
}
//...
return Ok("success");
}

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这是正确的效果，现在如果我覆盖这个SaveUser操作，并传递新的请求模型UserModelSex，我得到了一个错误，因为重写方法必须有相同的参数列表与父方法，所以我改变它像这样

public class User100Controller : ApiController
{
[HttpPost]
[SwaggerResponse(200, "success", typeof(UserModel))]
public virtual IHttpActionResult SaveUser([FromBody] JObject json)
{
var model = json.ToObject<UserModel>();
if (string.IsNullOrEmpty(model.name) || model.age == 0)
{
return Ok("error");
}
//...
return Ok("success");
}
}
public class User101Controller : User100Controller
{
[HttpPost]
[SwaggerResponse(200, "success", typeof(UserModelSex))]
public override IHttpActionResult SaveUser([FromBody] JObject json)
{
var model = json.ToObject<UserModelSex>();
if (string.IsNullOrEmpty(model.name) || model.age == 0 || model.sex == -1)
{
return Ok("error");
}
//...
return Ok("success");
}
}

swagger ui不识别jobobject

样图，请右键单击

如何显示UserModel和UserModeSex在请求模型中，如PIC 2

Answer 1

最后，我自己解决了。

• 添加SwaggerRequestModelAttribute.cs

[AttributeUsage(AttributeTargets.Method, AllowMultiple = false)]
public class SwaggerRequestModelAttribute : Attribute
{
public Type RequestModel { get; private set; }
public string ModelName { get; private set; }
public SwaggerRequestModelAttribute(Type requestModel)
{
RequestModel = requestModel;
ModelName = requestModel.Name;
}
}

• 然后，将[SwaggerRequestModel]标记到控制器

[HttpPost]
[SwaggerRequestModel(typeof(UserModel))]
public virtual IHttpActionResult SaveUser([FromBody] JObject json)
{
var model = json.ToObject<UserModel>();
if (string.IsNullOrEmpty(model.name) || model.age == 0)
{
return Ok("error");
}
//...
return Ok("success");
}

• 添加ModelInBodyOperationFilter.cs

public class ModelInBodyOperationFilter : IOperationFilter
{
public void Apply(Operation operation, SchemaRegistry schemaRegistry, ApiDescription apiDescription)
{
if (operation.parameters == null) operation.parameters = new List<Parameter>();
var attribute = apiDescription.GetControllerAndActionAttributes<SwaggerRequestModelAttribute>();
if (attribute.Any())
{
if (operation.parameters.Count > 0 && operation.parameters[0].schema.type == "object")
{
if (!schemaRegistry.Definitions.ContainsKey(attribute.First().ModelName))
schemaRegistry.GetOrRegister(attribute.First().RequestModel);
operation.parameters.RemoveAt(0);
operation.parameters.Add(new Parameter
{
name = "-",
@in = "body",
required = true,
schema = new Schema { @ref = $"#/definitions/{attribute.First().RequestModel.Namespace}.{attribute.First().ModelName}" }
});
}
}
}
}

• 也在SwaggerConfig.Register()
中

c.UseFullTypeNameInSchemaIds();
c.OperationFilter<ModelInBodyOperationFilter>();

• 最后，运行应用程序并预览

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