Csharp 如何在NET 8最小API中自动记录TypedResults.Stream响应

从swagger docs人开始:

ApiExplorer(Swashbakle构建在其上的ASP.NET核心元数据组件)默认情况下不呈现FileResult类型，因此您需要使用Products属性显式地告诉它:

[HttpGet("{fileName}")]
[Produces("application/octet-stream", Type = typeof(FileResult))]
public FileResult GetFile(string fileName)

PushStreamHttpResult也是文件结果，似乎同样的规则也适用于它，因此您可以使用Produces调用，例如:

app.MapGet("api/test", async Task<Results<PushStreamHttpResult, NotFound>> (bool fail) =>
    .WithSummary("Test Streaming results")
    .Produces(200, contentType: "text/plain" )

如果所有PushStreamHttpResult个都是纯文本，那么您可以try 使用操作Filter将其自动化:

// sample implementation, adjust for production
public sealed class PushStreamHttpResultFilter : IOperationFilter
{
    public void Apply(OpenApiOperation operation, OperationFilterContext context)
    {
        var returnType = context.MethodInfo.ReturnType;

        if (ContainsPushStream(returnType))
        {
            // adds plain text only if no 200 are defined
            operation.Responses.TryAdd("200", new OpenApiResponse
            {
                Content = new Dictionary<string, OpenApiMediaType>
                {
                    ["text/plain"] = new OpenApiMediaType()
                    {
                        Schema = new OpenApiSchema()
                    }
                }
            });
        }

        bool ContainsPushStream(Type type)
        {
            if (type == typeof(PushStreamHttpResult))
                return true;
            if (!type.IsConstructedGenericType)
                return false;
            if (type.IsConstructedGenericType && type.GetGenericTypeDefinition() == typeof(Task<>))
                return ContainsPushStream(type.GetGenericArguments()[0]);
            return type.GetGenericArguments().Any(t => t == typeof(PushStreamHttpResult));
        }
    }
}

并将其添加到过滤器中:

builder.Services.AddSwaggerGen(o =>
{
    o.OperationFilter<PushStreamHttpResultFilter>();
});