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up vote 4 down vote favorite

Was wondering what the best way is to pull data from an API (in JSON format). I have code in my controller, which calls an API which returns data.

I want to get the data onto my View so i can display it on a page. I have seen the most documented way by using jQuery/AJAX, but i dont really want the API url's to be made public.

I was thinking of passing an object created from the returned data. But in all honesty i am not sure how to do this!

The below code brings back the data for the products, per user. This works fine.

public static List<productDetails> GetUserProducts(string userid)
{
    //api/product/user/<Guid>
    var url = baseUrl + "/product/user/" + userid;

    var syncClient = new WebClient();
    var content = syncClient.DownloadString(url);

    List<productDetails> Products = (List<productDetails>)Newtonsoft.Json.JsonConvert.DeserializeObject(content, typeof(List<productDetails>));

    return Products;
}

And at present i am passing the returned data to the page using ViewBag. This does not work well if there is more than one product. I am passing this in the ActionResult for the view.

var p = GetUserProducts(userGuid);

foreach(var product in p)
{
    ViewBag.pId = product.Id;
    ViewBag.pName = product.FriendlyName;
    ViewBag.pSerial = product.SerialNumber;
    ViewBag.pbatt = product.Location.BatteryCharge + "%";

    ViewBag.devicehistory = "~/Location/History/" + product.Id;
}

Any ideas/examples would be much appreciated.

share|improve this question
    
Why don't you just use Model instead of ViewBag? – cem May 12 '14 at 8:48
    
I have a model set up, i am pretty new to MVC tbh. do you have an example? – thatuxguy May 12 '14 at 8:53
up vote 5 down vote accepted

Hope this can give you some idea on how it actually work

Some example of return as actionresult to view

Controller

public ActionResult something(string userGuid)
{
    var p = GetUserProducts(userGuid);
    return view(p); //you can return as partial view  (return PartialView("your partial view name", p));
}

View

@model IEnumerable<productDetails>


 foreach (var item in Model)
{
   @Html.DisplayFor(model => item.Id)
   //and so on
}

JsonResult

Some example of return as json to view

Controller

   [httpPost]
    public JsonResult something(string userGuid)
    {
        var p = GetUserProducts(userGuid);
        return Json(p, JsonRequestBehavior.AllowGet);
    }

call with ajax

$.post( "../something", {userGuid: "foo"}, function( data ) {
  console.log(data)
});
share|improve this answer
    
First part works a treat, just what i needed i think :D Cheers! – thatuxguy May 12 '14 at 9:27
    
what if i had a 2nd -- var p2 = GetUserOtherProducts(userGuid); for example? how would i pass that to the view? would that be using partial views? – thatuxguy May 12 '14 at 9:30
1  
If both is totally different model(should be different model), then you need to create a ViewModel, you can refer this stackoverflow.com/questions/16548376/… for more info – Se0ng11 May 12 '14 at 9:33
    
cheers i will take a look – thatuxguy May 12 '14 at 10:01
    
used this one too -- stackoverflow.com/questions/6937156/… Thanks for your help! :D – thatuxguy May 12 '14 at 10:07

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