c# - Swagger Swashbuckle 抽象类文档

Question

我有具有以下属性的 Web API 请求模型:

public List<Feature> Features { get; set; }

功能是一个抽象类。我将从中派生出许多类:

public abstract class Feature
{
    public string Title { get; set; }
}

public class ImageFeature : Feature
{
    public string ImageUrl { get; set; }
}

显然，Swashbuckle 仅识别 Feature 属性并相应地生成文档。我如何显式声明 Feature 类的可能实现，以便 Swashbuckle 生成正确的文档？有没有我可以使用的属性，比如:

[SwaggerResponseType(typeof(ImageFeature))]
[SwaggerResponseType(typeof(AnotherFeature))]
public abstract class Feature

Answer 1

看看这个: http://swashbuckletest.azurewebsites.net/swagger/ui/index#!/InheritanceTest/InheritanceTest_Get

下面是该 Controller 背后的代码: https://github.com/heldersepu/SwashbuckleTest/blob/911bf68e0cf6af3ee5d8278e6dd988eda8c4dc8d/Swagger_Test/Controllers/InheritanceTestController.cs

using System.Web.Http;

namespace Swagger_Test.Controllers
{
    public abstract class Feature
    {
        /// <summary>We all need a Title</summary>
        public string Title { get; set; }
    }

    public class ImageFeature : Feature
    {
        /// <summary>Here goes your image URL</summary>
        public string ImageUrl { get; set; }
    }

    public class InheritanceTestController : ApiController
    {
        public ImageFeature Get([FromUri]ImageFeature imageFeature)
        {
            return imageFeature;
        }

        public ImageFeature Post(ImageFeature imageFeature)
        {
            return imageFeature;
        }
    }
}