Restsharp posting file and model data to web api

WebApiConfig .cs

In order to add support for multipart/form-data we need to add below line to webapi.config

config.Formatters.XmlFormatter.SupportedMediaTypes.Add(new System.Net.Http.Headers.MediaTypeHeaderValue("multipart/form-data"));

using System;

using System.Collections.Generic;

using System.Linq;

using System.Web.Http;

namespace RestSharpEx

{

public static class WebApiConfig

{

public static void Register(HttpConfiguration config)

{

// Web API configuration and services

// Web API routes

config.MapHttpAttributeRoutes();

config.Formatters.XmlFormatter.SupportedMediaTypes.Add(new System.Net.Http.Headers.MediaTypeHeaderValue("multipart/form-data"));

config.Routes.MapHttpRoute(

name: "DefaultApi",

routeTemplate: "api/{controller}/{id}",

defaults: new { id = RouteParameter.Optional }

);

}

MyApiController.cs

using System;

using System.Collections.Generic;

using System.IO;

using System.Linq;

using System.Net;

using System.Net.Http;

using System.Web;

using System.Web.Http;

using Newtonsoft.Json;

using RestSharpEx.Models;

public class MyApiController : ApiController

{

[HttpPost, Route("api/MyApi/BulkUploadFile")]

public void BulkUploadFile(FileUploadModel model)

{

HttpResponseMessage result = null;

var httpRequest = HttpContext.Current.Request;

//Access your variable using

var me = JsonConvert.DeserializeObject<FileUploadModel>((HttpContext.Current.Request.Params["model"]).ToString());

if (httpRequest.Files.Count > 0)

{

var docfiles = new List<string>();

foreach (string file in httpRequest.Files)

{

var postedFile = httpRequest.Files[file];

var filePath = HttpContext.Current.Server.MapPath("~/" + postedFile.FileName);

postedFile.SaveAs(filePath);

docfiles.Add(filePath);

}

result = Request.CreateResponse(HttpStatusCode.Created, docfiles);

}

Below line will be used to bring information of posted model as web api dont support it. Cant be sent as a body parameter.

//Access your variable using

var me = JsonConvert.DeserializeObject<FileUploadModel>((HttpContext.Current.Request.Params["model"]).ToString());

Mvc controller action

[HttpPost]

public ActionResult PostWithFile(FileUploadModel model)

{

if (Request.Files.Count > 0)

{

var file = Request.Files[0];

if (file != null && file.ContentLength > 0)

{

var fileName = Path.GetFileName(file.FileName);

var path = Path.Combine(Server.MapPath("~/Images/"), fileName);

file.SaveAs(path);

var apiUrl = @"http://localhost:8076/api/MyApi/BulkUploadFile";

var client = new RestClient(apiUrl);

//if (!string.IsNullOrWhiteSpace(userName))

//{

// client.Authenticator = new HttpBasicAuthenticator(userName, password);

//}

client.ClearHandlers();

//var jsonDeserializer = new JsonDeserializer();

//client.AddHandler("application/json", jsonDeserializer);

var request = new RestRequest(Method.POST);

request.AddHeader("Accept", "application/json");

request.Parameters.Clear();

request.AddParameter("model", JsonConvert.SerializeObject(model));

request.AddHeader("Content-Type", "multipart/form-data");

request.AddFile("content", path);

var result = client.Execute(request).Content;

var finalResult = JsonConvert.DeserializeObject<FileUploadModel>(result);

}

return View(model);

}

Below line is important

request.AddParameter("model", JsonConvert.SerializeObject(model));

request.AddHeader("Content-Type", "multipart/form-data");

request.AddFile("content", path);

As we are defining it as multipart/form-data and then adding model as a parameter and also adding a file to request using path.

View.cshtml

@model RestSharpEx.Models.FileUploadModel

ViewBag.Title = "PostWithFile";

}

<h2>PostWithFile</h2>

@*@using (Html.BeginForm())*@

@using (Html.BeginForm("PostWithFile", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))

{

@Html.AntiForgeryToken()

<h4>FileUploadModel</h4>

<hr />

@Html.ValidationSummary(true, "", new { @class = "text-danger" })

@Html.LabelFor(model => model.FileId, htmlAttributes: new { @class = "control-label col-md-2" })

@Html.EditorFor(model => model.FileId, new { htmlAttributes = new { @class = "form-control" } })

@Html.ValidationMessageFor(model => model.FileId, "", new { @class = "text-danger" })

</div>

@Html.LabelFor(model => model.FileDescription, htmlAttributes: new { @class = "control-label col-md-2" })

@Html.EditorFor(model => model.FileDescription, new { htmlAttributes = new { @class = "form-control" } })

@Html.ValidationMessageFor(model => model.FileDescription, "", new { @class = "text-danger" })

</div>

@Html.LabelFor(model => model.FileDescription, htmlAttributes: new { @class = "control-label col-md-2" })

</div>

</div>

}

<div>

@Html.ActionLink("Back to List", "Index")

</div>

@section Scripts {

@Scripts.Render("~/bundles/jqueryval")

}

Note : We need to use @using (Html.BeginForm("PostWithFile", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))

in begin form in order to send to model with file upload data.

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Restsharp posting file and model data to web api