Swagger EF Core 3.1 Delta<Object> 请求正文 - 显示对象架构而不是 Odata.Delta

Question

我最近开始在我的 EF Core 项目中使用 Swagger/Swashbuckle。

当我将补丁 API 添加到文档时，请求正文的示例值/架构显示的是 Delta 的详细信息，而不是我的对象。由于它是需要发布的对象属性的子集，有没有办法显示我的对象架构？

大摇大摆的表现

我想要什么（屏幕截图来自 Post API 调用）

swagger 和 API 的签名如下。

/// <summary>
/// Updates the provided Absence Reason.
/// </summary>
/// <remarks>
/// Runs bespoke Patch/Validation logic.
/// Updates the Absence Reason then returns the record.
/// </remarks>
/// <param name="key" required="true">Id of the Absence Reason being updated.</param>
/// <param name="delta" required="true">A delta of the updated Absence Reason record.</param>
/// <returns>Returns the created Absence Reason</returns>
/// <response code="204">Returns the updated Absence Reason.</response>
/// <response code="400">Invalid Absence Reason Record, Missing Row Version etc.</response>
/// <response code="404">Absence Reason not found.</response>
/// <response code="412">Record has been updated since the version provided.</response>
[ApiExplorerSettings]
[HttpPatch("odata/[controller]({key})")]
[Produces("application/json")]
[ProducesResponseType(typeof(AbsenceReason), StatusCodes.Status204NoContent)]
[ProducesResponseType(400)]
[ProducesResponseType(404)]
[ProducesResponseType(412)]
[Authorize(Policy = AuthPolicyNames.GOIAdminRole)]
public IActionResult Patch([FromODataUri] long key, Delta<AbsenceReason> delta)

编辑以添加其他信息以响应 AlFranco

Watch on Reference 的屏幕截图

您代码中的 _assemblyName 是 “Microsoft.AspNetCore.OData，版本=7.4.1.0，文化=中性，PublicKeyToken=31bf3856ad364e35”

VS 文档创建的精简后的 XML 文件是：

        <member name="M:ESRGOI.Controllers.AbsenceReasonsController.Patch(System.Int64,Microsoft.AspNet.OData.Delta{ESRGOI.Models.AbsenceReason})">
            <summary>
            Updates the provided Absence Reason.
            </summary>
            <param name="key" required="true">Id of the Absence Reason being updated.</param>
            <param name="delta" required="true">A delta of the updated Absence Reason record.</param>
        </member>

编辑 2 好的，我已经根据 AlFranco 的回复想出了一个解决方案。

应该注意在 Swagger 中我的模型不包含命名空间。我最初尝试包含命名空间，但收到以下错误： /components/schemas/ESRGOI.Models.AbsenceReason 在文档中不存在

public class DeltaOperationFilter : IOperationFilter
{
  private const string _deltaParam = "Delta";

  public void Apply(OpenApiOperation operation, OperationFilterContext context)
  {
    if (operation.RequestBody == null) return;

    var deltaTypes =
        operation.RequestBody
            .Content
            .Where(x => x.Value.Schema.Reference.Id.EndsWith(_deltaParam));

    foreach (var (_, value) in deltaTypes)
    {
      var schema = value.Schema;
      string model = schema.Reference.Id.Substring(0, schema.Reference.Id.Length - _deltaParam.Length);
      schema.Reference.Id = model;
    }
  }
}

Answer 1

好的，为此你需要创建一个IOperationFilter

这对我有用：

    public class DeltaOperationFilter : IOperationFilter
    {
        private const string DeltaWrapper = "Microsoft.AspNet.OData.Delta`1";
        private readonly string _assemblyName = typeof(Microsoft.AspNet.OData.Delta).Assembly.FullName;

        public void Apply(OpenApiOperation operation, OperationFilterContext context)
        {                                                               
            if (operation.RequestBody == null) return;
            
            var deltaTypes =
                operation.RequestBody
                    .Content
                    .Where(x => x.Value.Schema.Reference.Id.StartsWith(DeltaWrapper));

            foreach (var (_, value) in deltaTypes)
            {
                var schema = value.Schema;
                var deltaType = Type.GetType(schema.Reference.Id+ ", " + _assemblyName);
                var deltaArgument = deltaType?.GetGenericArguments().First();
                schema.Reference.Id = deltaArgument?.FullName ?? schema.Reference.Id;
            }
        }
    }

之后你唯一需要做的就是在 swaggergen 中注册

// Assuming you are in Startup.cs ConfigureServices
services.AddSwaggerGen(c => c.OperationFilter<DeltaOperationFilter>());

顺便问一下，您能否在您的实体上正确使用 Delta<T> 应用？我将使用带有某些端点的补丁，但想使用Delta<DTO>，然后对其进行转换并将其应用于Delta<Entity>，这实际上会持续存在。我对您将采用的方法特别感兴趣:)

谢谢！