Call UrlHelper in models in ASP.NET MVC

Question

I need to generate some URLs in a model in ASP.NET MVC. I'd like to call something like UrlHelper.Action() which uses the routes to generate the URL. I don't mind filling the usual blanks, like the hostname, scheme and so on.

Is there any method I can call for that? Is there a way to construct an UrlHelper?

Answer 1

Helpful tip, in any ASP.NET application, you can get a reference of the current HttpContext

HttpContext.Current

which is derived from System.Web. Therefore, the following will work anywhere in an ASP.NET MVC application:

UrlHelper url = new UrlHelper(HttpContext.Current.Request.RequestContext);
url.Action("ContactUs"); // Will output the proper link according to routing info

Example:

public class MyModel
{
    public int ID { get; private set; }
    public string Link
    {
        get
        {
            UrlHelper url = new UrlHelper(HttpContext.Current.Request.RequestContext);
            return url.Action("ViewAction", "MyModelController", new { id = this.ID });
        }
    }

    public MyModel(int id)
    {
        this.ID = id;
    }
}

Calling the Link property on a created MyModel object will return the valid Url to view the Model based on the routing in Global.asax

Answer 2

I like Omar's answer but that's not working for me. Just for the record this is the solution I'm using now:

var httpContext = HttpContext.Current;

if (httpContext == null) {
  var request = new HttpRequest("/", "http://example.com", "");
  var response = new HttpResponse(new StringWriter());
  httpContext = new HttpContext(request, response);
}

var httpContextBase = new HttpContextWrapper(httpContext);
var routeData = new RouteData();
var requestContext = new RequestContext(httpContextBase, routeData);

return new UrlHelper(requestContext);

Answer 3

A UrlHelper can be constructed from within a Controller action with the following:

 var url = new UrlHelper(this.ControllerContext.RequestContext);
 url.Action(...);

Outside of a controller, a UrlHelper can be constructed by creating a RequestContext from the RouteTable.Routes RouteData.

HttpContextWrapper httpContextWrapper = new HttpContextWrapper(System.Web.HttpContext.Current);
UrlHelper urlHelper = new UrlHelper(new RequestContext(httpContextWrapper, RouteTable.Routes.GetRouteData(httpContextWrapper)));

(Based on Brian's answer, with a minor code correction added.)

Answer 4

Yes, you can instantiate it. You can do something like:

var ctx = new HttpContextWrapper(HttpContext.Current);
UrlHelper helper = new UrlHelper(
   new RequestContext(ctx,
   RouteTable.Routes.GetRouteData(ctx));

RouteTable.Routes is a static property, so you should be OK there; to get a HttpContextBase reference, HttpContextWrapper takes a reference to HttpContext, and HttpContext delivers that.

Answer 5

After trying all the other answers, I ended up with

$"/api/Things/Action/{id}"

Haters gonna hate ¯\_(ツ)_/¯

Answer 6

I was trying to do something similar from within a page (outside of a controller).

UrlHelper did not allow me to construct it as easily as Pablos answer, but then I remembered a old trick to effective do the same thing:

string ResolveUrl(string pathWithTilde)

Answer 7

Previous responses didn't help me. My approach has been create an action in my Home controller with the same functionality that UrlHelper.

    [HttpPost]
    [Route("format-url")]
    public ActionResult FormatUrl()
    {
        string action = null;
        string controller = null;
        string protocol = null;
        dynamic parameters = null;

        foreach (var key in this.Request.Form.AllKeys)
        {
            var value = this.Request.Form[key];
            if (key.Similar("action"))
            {
                action = value;
            }
            else if (key.Similar("controller"))
            {
                controller = value;
            }
            else if (key.Similar("protocol"))
            {
                protocol = value;
            }
            else if (key.Similar("parameters"))
            {
                JObject jObject = JsonConvert.DeserializeObject<dynamic>(value);

                var dict = new Dictionary<string, object>();
                foreach (var item in jObject)
                {
                    dict[item.Key] = item.Value;
                }

                parameters = AnonymousType.FromDictToAnonymousObj(dict);
            }
        }

        if (string.IsNullOrEmpty(action))
        {
            return new ContentResult { Content = string.Empty };
        }

        int flag = 1;
        if (!string.IsNullOrEmpty(controller))
        {
            flag |= 2;
        }

        if (!string.IsNullOrEmpty(protocol))
        {
            flag |= 4;
        }

        if (parameters != null)
        {
            flag |= 8;
        }

        var url = string.Empty;
        switch (flag)
        {
            case 1: url = this.Url.Action(action); break;
            case 3: url = this.Url.Action(action, controller); break;
            case 7: url = this.Url.Action(action, controller, protocol); break;
            case 9: url = this.Url.Action(action, parameters); break;
            case 11: url = this.Url.Action(action, controller, parameters); break;
            case 15: url = this.Url.Action(action, controller, parameters, protocol); break;
        }

        return new ContentResult { Content = url };
    }

Been an action, you can request it from anywhere, even inside the Hub:

            var postData = "action=your-action&controller=your-controller";

            // Add, for example, an id parameter of type integer
            var json = "{\"id\":3}";
            postData += $"&parameters={json}";

            var data = Encoding.ASCII.GetBytes(postData);

#if DEBUG
            var url = $"https://localhost:44301/format-url";
#else
            var url = $"https://your-domain-name/format-url";
#endif

            var request = (HttpWebRequest)WebRequest.Create(url);
            request.Method = "POST";
            request.ContentType = "application/text/plain";
            request.ContentLength = data.Length;

            using (var stream = request.GetRequestStream())
            {
                stream.Write(data, 0, data.Length);
            }

            var response = (HttpWebResponse)request.GetResponse();
            var link = new StreamReader(stream: response.GetResponseStream()).ReadToEnd();

You can get source code of AnonymousType here.

Answer 8

I think what you're looking for is this:

Url.Action("ActionName", "ControllerName");