如何手动添加到 swagger 的 schemas 部分？

Question

我正在使用 Swashbuckle.AspNetCore 并将此模式部分开箱即用：

我的响应模式看起来是空的，但实际上我返回了子类。我怎样才能将它们添加到此部分？

我有这个请求模型：

using System;
using System.Collections.Generic;
using System.ComponentModel.DataAnnotations;
using System.Linq;
using System.Runtime.Serialization;
using Newtonsoft.Json;
using static Assignment_1.AppGlobals;

namespace Assignment_1
{
    public class ExecuteMoveRequest: IValidatableObject, ICloneable
    {
        public int? Move { get; set; }

        [Required]
        public BoardSymbol AzurePlayerSymbol { get; set; }

        [Required]
        public BoardSymbol HumanPlayerSymbol { get; set; }

        [MinLength(9)]
        [MaxLength(9)]
        public BoardSymbol[] GameBoard { get; set; }

        public IEnumerable<ValidationResult> Validate(ValidationContext validationContext)
        {
//...

我将其用于我的响应模型：

using System;
using System.ComponentModel.DataAnnotations;
using System.Runtime.Serialization;
using static Assignment_1.AppGlobals;

namespace Assignment_1
{
    public interface ExecuteMoveResponse
    {
    }

    [KnownType(typeof(ValidExecuteMoveResponse))]
    public class ValidExecuteMoveResponse: ExecuteMoveRequest, ExecuteMoveResponse
    {
        public string type { get; set; }
    }

    [KnownType(typeof(InProgressExecuteMoveResponse))]
    public class InProgressExecuteMoveResponse : ValidExecuteMoveResponse
    {
    }

    [KnownType(typeof(WonExecuteMoveResponse))]
    public class WonExecuteMoveResponse : ValidExecuteMoveResponse
    {
        [Required]
        public BoardSymbol Winner { get; set; }
        [Required]
        public int[] WinPositions { get; set; }
    }
}

Answer 1

首先尝试将这些类分离到不同的文件中。第二次尝试在您的课程之前添加以下评论：

/// <summary>
/// Description for Your class
/// </summary>
[KnownType(typeof(InProgressExecuteMoveResponse))]
public class InProgressExecuteMoveResponse : ValidExecuteMoveResponse
{
    ...
}

在某些情况下，要显示摘要，您必须在 swashbuckle 配置中启用它。还尝试以这种方式配置 Swashbuckle：

  app.UseSwaggerUi3WithApiExplorer(settings =>
        {
            settings.GeneratorSettings.DefaultEnumHandling = EnumHandling.String;
            settings.GeneratorSettings.AllowReferencesWithProperties = true;
...