我可以指定 Web-Api 方法需要 Swashbuckle/Swagger 生成的页面/json 中的文件吗

Question

I got a Web-Api with the following simplified method for obtaining a file:

[HttpPut("Test/{id}")]
public IActionResult PutTest(string id)
{
    Stream file = Request.Body;
    //rest of method
    return StatusCode(201);
}

It works fine, although in the SwaggerUI page that gets created there's no mention of the method expecting a file in the body. Is there a way to specify the method expects a file in the generated SwaggerUI page?

I can simulate the input using the following piece of code:

var content = new StreamContent(new FileStream("C:\temp\test.txt", FileMode.Open));

using (var client = new HttpClient())
using (var response = client.PutAsync(url, content))
{
    var result = response.Result;
    //more code   
}

Answer 1

The file upload is not documented because it's not a param in the action method. Swagger has no way of knowing what you're doing inside the action itself. Attempting to handle a file upload this way is bad practice anyways. You can use IFormFile , but that only works if your request body is encoded as multipart/form-data . If you're dealing with JSON or basically anything that qualifies as FromBody , then you need to bind to a byte[] :

[HttpPut("Test/{id}")]
public IActionResult PutTest(string id, byte[] file)
{
    //rest of method
    return StatusCode(201);
}

Now, the automatic FromBody applied by the [ApiController] attribute only works with class types, so I'm not sure off the top of my head whether it will apply to a byte[] . If not, just do:

public IActionResult PutTest(string id, [FromBody]byte[] file)