Calling Web API from MVC controller

Question

I have a WebAPI Controller within my MVC5 project solution. The WebAPI has a method which returns all files in a specific folder as a Json list:

[{"name":"file1.zip", "path":"c:\\"}, {...}]

From my HomeController I want to call this Method, convert the Json response to List<QDocument> and return this list to a Razor view. This list might be empty: [] if there are no files in the folder.

This is the APIController:

public class DocumentsController : ApiController {     #region Methods     /// <summary>     /// Get all files in the repository as Json.     /// </summary>     /// <returns>Json representation of QDocumentRecord.</returns>     public HttpResponseMessage GetAllRecords()     {       // All code to find the files are here and is working perfectly...           return new HttpResponseMessage()          {              Content = new StringContent(JsonConvert.SerializeObject(listOfFiles), Encoding.UTF8, "application/json")          };     }                } 

Here is my HomeController:

public class HomeController : Controller {      public Index()      {       // I want to call APi GetAllFiles and put the result to variable:       var files = JsonConvert.DeserializeObject<List<QDocumentRecord>>(API return Json);       }  } 

Finally this is the model in case you need it:

public class QDocumentRecord {       public string id {get; set;}       public string path {get; set;}    ..... } 

So how can I make this call?

Answer 1

From my HomeController I want to call this Method and convert Json response to List

No you don't. You really don't want to add the overhead of an HTTP call and (de)serialization when the code is within reach. It's even in the same assembly!

Your ApiController goes against (my preferred) convention anyway. Let it return a concrete type:

public IEnumerable<QDocumentRecord> GetAllRecords() {     listOfFiles = ...     return listOfFiles; } 

If you don't want that and you're absolutely sure you need to return HttpResponseMessage, then still there's absolutely no need to bother with calling JsonConvert.SerializeObject() yourself:

return Request.CreateResponse<List<QDocumentRecord>>(HttpStatusCode.OK, listOfFiles); 

Then again, you don't want business logic in a controller, so you extract that into a class that does the work for you:

public class FileListGetter {     public IEnumerable<QDocumentRecord> GetAllRecords()     {         listOfFiles = ...         return listOfFiles;     } } 

Either way, then you can call this class or the ApiController directly from your MVC controller:

public class HomeController : Controller {     public ActionResult Index()     {         var listOfFiles = new DocumentsController().GetAllRecords();         // OR         var listOfFiles = new FileListGetter().GetAllRecords();          return View(listOfFiles);     } } 

But if you really, really must do an HTTP request, you can use HttpWebRequest, WebClient, HttpClient or RestSharp, for all of which plenty of tutorials exist.

Answer 2

Its very late here but thought to share below code. If we have our WebApi as a different project altogether in the same solution then we can call the same from MVC controller like below

public class ProductsController : Controller     {         // GET: Products         public async Task<ActionResult> Index()         {             string apiUrl = "http://localhost:58764/api/values";              using (HttpClient client=new HttpClient())             {                 client.BaseAddress = new Uri(apiUrl);                 client.DefaultRequestHeaders.Accept.Clear();                 client.DefaultRequestHeaders.Accept.Add(new System.Net.Http.Headers.MediaTypeWithQualityHeaderValue("application/json"));                  HttpResponseMessage response = await client.GetAsync(apiUrl);                 if (response.IsSuccessStatusCode)                 {                     var data = await response.Content.ReadAsStringAsync();                     var table = Newtonsoft.Json.JsonConvert.DeserializeObject<System.Data.DataTable>(data);                  }               }             return View();          }     }