如何大摇大摆地记录一个直接读取 Request.Body 的 ASP.NET 核心操作

Question

我有一个 Controller 操作方法，它直接（而不是使用文件）读取Request.Body用于流式传输和其他目的。 问题是没有 model 绑定，因此 Swagger 没有记录合同。 例如：

[HttpPost("upload")]
[DisableFormValueModelBinding]
public async Task<IActionResult> UploadAsync()
{
  // Read from Request.Body manually, expecting content type to be multipart/*
  return Ok();
}

加载Swagger UI时，无法上传文件等。

有没有办法通过 ASP.NET 核心中的属性来支持这一点？

Answer 1

API：

 [HttpPost]  
    public async Task<IActionResult> Post(  
        [FromForm(Name = "myFile")]IFormFile myFile)  
    {  
            using (var fileContentStream = new MemoryStream())  
            {  
                await myFile.CopyToAsync(fileContentStream);  
                await System.IO.File.WriteAllBytesAsync(Path.Combine(folderPath, myFile.FileName), fileContentStream.ToArray());  
            }  
            return CreatedAtRoute(routeName: "myFile", routeValues: new { filename = myFile.FileName }, value: null); ;  
    }  

操作过滤器

public class SwaggerFileOperationFilter : IOperationFilter  
{  
    public void Apply(Operation operation, OperationFilterContext context)  
    {  
        if (operation.OperationId == "Post")  
        {  
            operation.Parameters = new List<IParameter>  
            {  
                new NonBodyParameter  
                {  
                    Name = "myFile",  
                    Required = true,  
                    Type = "file",  
                    In = "formData"  
                }  
            };  
        }  
    }  
}  

启动-ConfigureServices

services.AddSwaggerGen(  
    options =>  
    {  
        options.SwaggerDoc("v1", new Swashbuckle.AspNetCore.Swagger.Info { Title = "My API", Version = "v1" });  

        options.OperationFilter<SwaggerFileOperationFilter>();  
    });  

swagger UI 中的结果：

来源是： 在此处输入链接描述