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copied!<p>When you do <code>return Json(...)</code> you are specifically telling MVC <em>not to use a view</em>, and to serve serialized JSON data. Your browser opens a download dialog because it doesn't know what to do with this data.</p> <p>If you instead want to return a view, just do <code>return View(...)</code> like you normally would:</p> <pre><code>var dictionary = listLocation.ToDictionary(x => x.label, x => x.value); return View(new { Values = listLocation }); </code></pre> <p>Then in your view, simply encode your data as JSON and assign it to a JavaScript variable:</p> <pre><code><script> var values = @Html.Raw(Json.Encode(Model.Values)); </script> </code></pre> <hr> <p><strong>EDIT</strong></p> <p>Here is a bit more complete sample. Since I don't have enough context from you, this sample will assume a controller <code>Foo</code>, an action <code>Bar</code>, and a view model <code>FooBarModel</code>. Additionally, the list of locations is hardcoded:</p> <p><strong>Controllers/FooController.cs</strong></p> <pre><code>public class FooController : Controller { public ActionResult Bar() { var locations = new[] { new SelectListItem { Value = "US", Text = "United States" }, new SelectListItem { Value = "CA", Text = "Canada" }, new SelectListItem { Value = "MX", Text = "Mexico" }, }; var model = new FooBarModel { Locations = locations, }; return View(model); } } </code></pre> <p><strong>Models/FooBarModel.cs</strong></p> <pre><code>public class FooBarModel { public IEnumerable<SelectListItem> Locations { get; set; } } </code></pre> <p><strong>Views/Foo/Bar.cshtml</strong></p> <pre><code>@model MyApp.Models.FooBarModel <script> var locations = @Html.Raw(Json.Encode(Model.Locations)); </script> </code></pre> <p>By the looks of your error message, it seems like you are mixing incompatible types (i.e. <code>Ported_LI.Models.Locatio‌n</code> and <code>MyApp.Models.Location</code>) so, to recap, make sure the type sent from the controller action side match what is received from the view. For this sample in particular, <code>new FooBarModel</code> in the controller matches <code>@model MyApp.Models.FooBarModel</code> in the view.</p>

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