Asp.net-mvc – Html.ActionLink as a button or an image, not a link

I think what you want is this:

ASP.NET MVC1

Html.ActionLink(article.Title, 
                "Login",  // <-- Controller Name.
                "Item",   // <-- ActionMethod
                new { id = article.ArticleID }, // <-- Route arguments.
                null  // <-- htmlArguments .. which are none. You need this value
                      //     otherwise you call the WRONG method ...
                      //     (refer to comments, below).
                )

This uses the following method ActionLink signature:

public static string ActionLink(this HtmlHelper htmlHelper, 
                                string linkText,
                                string controllerName,
                                string actionName,
                                object values, 
                                object htmlAttributes)

ASP.NET MVC2

two arguments have been switched around

Html.ActionLink(article.Title, 
                "Item",   // <-- ActionMethod
                "Login",  // <-- Controller Name.
                new { id = article.ArticleID }, // <-- Route arguments.
                null  // <-- htmlArguments .. which are none. You need this value
                      //     otherwise you call the WRONG method ...
                      //     (refer to comments, below).
                )

This uses the following method ActionLink signature:

public static string ActionLink(this HtmlHelper htmlHelper, 
                                string linkText,
                                string actionName,
                                string controllerName,
                                object values, 
                                object htmlAttributes)

ASP.NET MVC3+

arguments are in the same order as MVC2, however the id value is no longer required:

Html.ActionLink(article.Title, 
                "Item",   // <-- ActionMethod
                "Login",  // <-- Controller Name.
                new { article.ArticleID }, // <-- Route arguments.
                null  // <-- htmlArguments .. which are none. You need this value
                      //     otherwise you call the WRONG method ...
                      //     (refer to comments, below).
                )

This avoids hard-coding any routing logic into the link.

 <a href="/Item/Login/5">Title</a> 

This will give you the following html output, assuming:

1. article.Title = "Title"
2. article.ArticleID = 5
3. you still have the following route defined

. .

routes.MapRoute(
    "Default",     // Route name
    "{controller}/{action}/{id}",                           // URL with parameters
    new { controller = "Home", action = "Index", id = "" }  // Parameter defaults
);

Here's what I came up with, and it's working for me. I added the following method(s) to my controller base class. (You can always make these static methods somewhere else that accept a controller as a parameter I suppose)

MVC2 .ascx style

protected string RenderViewToString<T>(string viewPath, T model) {
  ViewData.Model = model;
  using (var writer = new StringWriter()) {
    var view = new WebFormView(ControllerContext, viewPath);
    var vdd = new ViewDataDictionary<T>(model);
    var viewCxt = new ViewContext(ControllerContext, view, vdd,
                                new TempDataDictionary(), writer);
    viewCxt.View.Render(viewCxt, writer);
    return writer.ToString();
  }
}

Razor .cshtml style

public string RenderRazorViewToString(string viewName, object model)
{
  ViewData.Model = model;
  using (var sw = new StringWriter())
  {
    var viewResult = ViewEngines.Engines.FindPartialView(ControllerContext,
                                                             viewName);
    var viewContext = new ViewContext(ControllerContext, viewResult.View,
                                 ViewData, TempData, sw);
    viewResult.View.Render(viewContext, sw);
    viewResult.ViewEngine.ReleaseView(ControllerContext, viewResult.View);
    return sw.GetStringBuilder().ToString();
  }
}

Edit: added Razor code.