如何在swagger codegen中包含列表属性?

swagger忽略ComplexItems属性。我需要为我的客户端应用程序生成axios模型。

public class DummyModel
{
    [Key]
    public Guid Id { get; set; }

    [Required]
    public string Name { get; set; }

    // This is your complex list
    public List<ComplexItem> ComplexItems { get; set; } = new List<ComplexItem>();
}

public class ComplexItem
{
    [Key]
    public Guid Id { get; set; }
    public int Property1 { get; set; }
    public string Property2 { get; set; }
    public DateTime Property3 { get; set; }
}

在Api上下文中,我添加了关系。

modelBuilder.Entity<DummyModel>()
        .HasMany(dm => dm.ComplexItems)
        .WithOne()  // or WithMany() based on your needs
        .HasForeignKey("ForeignKeyColumnIfAny");  

在我添加的控制器中包括:

return await _context.DummyModels
               .Include(dm => dm.ComplexItems)
               .ToListAsync();

但在没有列表属性的情况下,模型看起来是这样的:

export interface DummyModel {
   /**
    * 
    * @type {string}
    * @memberof DummyModel
    */
   id?: string;
   /**
    * 
    * @type {string}
    * @memberof DummyModel
    */
   name: string;
}

使现代化

程序.cs

    app.UseSwagger();
    app.UseSwaggerUI(c =>
    {
        c.SwaggerEndpoint("/swagger/v1/swagger.json", "My API V1");
    });
    ....
    builder.Services.AddSwaggerGen(c =>
{
    c.SwaggerDoc("v1", new OpenApiInfo { Title = "My API", Version = "v1" });
    c.SchemaFilter<ExcludeNullableSchemaFilter>();
});

DummyController

 [HttpPost("DummyModelTest")]
        public List<DummyModel> Post(DummyModel dummyModel)
        {

            List<DummyModel> list = new List<DummyModel> { dummyModel };
            return list;
        }