Error sending file via Postasjsonasync method in Asp.Net MVC

Asked

Viewed 524 times

3

In my Controller of a project Asp.net MVC I get from View a picture like HttpPostedFileBase and need to send to Web Api using PostAsJsonAsync:

var response = await client.PostAsJsonAsync("api/image", image);

The image is coming normally, but when arriving at the above line appears this error:

"Error getting value from 'ReadTimeout' on 'System.Web.HttpInputStream'."

Note: sending "image.Inputstream" from the same error.

How to solve this problem? Thank you.

Update: the Web Api I implemented based on this response: How to Save and Return Images with Web Api?

c# asp.net-mvc asp.net-mvc-5 asp.net-web-api

  • If you have the code, ie the methods? you can post?

  • You don’t have to use image.Inputstream?

  • @f. fujihara I tried to send the image.Inputstream and gives the same error.

  • the method it receives and send to the webapi (both if possible)

  • @Harrypotter made the based on that answer, but I haven’t tested yet pq I have this problem that prevents the image from getting there: http://answall.com/a/19461/8356

  • What you want to do with the image?

  • @f. fujihara have an Asp.net mvc project that in the controller gets this image from View and I want to send it to another Web Api project. In this project I want to receive the image and save it in a directory. In short: save the image and be able to return it, using the Asp.net MVC together Web Api (separate projects that are in the same Solution). What I’m trying to do is this: http://answall.com/questions/19454/como-salvar-e-retornar-imagens-com-web-api

  • You receive an Httppostedfilebase parameter in the Controller, and if in Business Voce you receive a Stream? parameter (image.Inputstream). It would be easier for you to manipulate and send to another directory.

  • 1

    Well I posted an example of how I would, I hope it helps, @Carlosgeovanio

  • 1

    @Harrypotter I’ll test that answer of yours, thank you very much.

  • OK @Carlosgeovanio

  • It worked @Carlosgeovanio ?

  • gave yes @Harrypotter. Thank you very much.

  • if you can close the question to serve as a model for others

Show 9 more comments

1 answer

2


Example:

This class will serve as the basis for transporting the image with its name and the byte array.

public class Transporte
{
    public string Name { get; set; }
    public byte[] Imagem { get; set; }
}

Action Methods (Web MVC)

[HttpGet]
public ActionResult EnviarImagem()
{
    return View();
}

[HttpPost]
public void EnviarImagem(HttpPostedFileBase imagem)
{
    int tam = (int)imagem.InputStream.Length;
    byte[] imgByte = new byte[tam];
    imagem.InputStream.Read(imgByte, 0, tam);

    HttpClient web = new HttpClient();
    web.BaseAddress = new Uri("http://localhost:17527/");
    HttpResponseMessage response = web.PostAsJsonAsync("api/GravarImagem", new Transporte() { Imagem = imgByte, Name = imagem.FileName }).Result;            
}

View referring to these methods

@{
    Layout = null;
}

<!DOCTYPE html>

<html>
<head>
    <meta name="viewport" content="width=device-width" />
    <title>EnviarImagem</title>
</head>
<body>
    <div> 
        <form method="post" action="/Home/EnviarImagem" enctype="multipart/form-data">
            <input type="file" name="imagem" id="imagem" />
            <button>Enviar</button>
        </form>
    </div>
</body>
</html>

Web Api

public class RecebeController : ApiController
{
    [HttpPost]
    [Route("api/GravarImagem")]
    public HttpResponseMessage GravarImagem(Transporte transporte)
    {
        File.WriteAllBytes(HttpContext.Current.Server.MapPath("~/Fotos/") + transporte.Name, transporte.Imagem);
        return Request.CreateResponse();
    }
}

Debug:

inserir a descrição da imagem aqui

Browser other questions tagged c# asp.net-mvc asp.net-mvc-5 asp.net-web-api

You are not signed in. Login or sign up in order to post.