HttpWebRequest - ASP .NET MVC 3 Passing Session State

Question

I need to acquire HTML markup of a controller/action in order to generate PDF. What I have done is:

    public ActionResult Index()
    {
        Session["Message"] = "SESSION-MESSAGE";

        String URL = "http://localhost:7401/Home/SuperComplex";

        HttpWebRequest req = (HttpWebRequest)WebRequest.Create(URL);
        req.CookieContainer = new CookieContainer();

        for (int i = 0; i <= this.Request.Cookies.Count - 1; i++)
            req.CookieContainer.Add(
                new System.Net.Cookie(
                    name: this.Request.Cookies[i].Name,
                    value: Request.Cookies[i].Value,
                    path: Request.Cookies[i].Path, domain: this.HttpContext.Request.Url.Host)
                );

        using (var r = req.GetResponse())
        {
            using (var s = new StreamReader(r.GetResponseStream()))
            {
                var htmlToPrint = s.ReadToEnd();
                Response.Write("<h1>" + htmlToPrint + "</h1>");
            }
        }

        return View();
    }

Considering above said situation, in SuperComplex session, I should have the Session["Message"]. But for some strange reason, it does not go there.

I have checked Session.SessionId - in both cases it is same.

Also, on second or third request, request timesout!

Again: http://localhost:7401/(S(SESSION_ID))/Home/About

If requested in other browser: session hijack does happen - but WebRequest dies! :(

Help - anyone?

Is it specific to ASP.NET MVC 4 (which is in Beta now)? If so, change the tag to `asp.net-mvc-4`. If not, change your question's title. — Ofer Zelig, Feb 20 '12 at 19:05
You are not using a webrequest to get the html from your own website aren't you? There are better means for this. — usr, Feb 20 '12 at 20:01
@usr: yes, I am trying to do so.. :(.. any specific mean you would like to mention? — effkay, Feb 20 '12 at 20:08

score 2 · Accepted Answer · answered Feb 29 '12 at 08:02

Store your HTML in a Partial View and then use a Helper function to parse it into a string.

// usage
/*
 * http://stackoverflow.com/questions/4344533/asp-net-mvc-razor-how-to-render-a-razor-partial-views-html-inside-the-controll
 * 
    var model = _repository.Find(x => x.PropertyID > 3).FirstOrDefault();
    var test = this.RenderViewToString("DataModel", model);
    return Content(test);
 */

public static string RenderPartialToString<T>(this ControllerBase controller, string partialName, T model)
{
    var vd = new ViewDataDictionary(controller.ViewData);
    var vp = new ViewPage
    {
        ViewData = vd,
        ViewContext = new ViewContext(),
        Url = new UrlHelper(controller.ControllerContext.RequestContext)
    };

    ViewEngineResult result = ViewEngines
                              .Engines
                              .FindPartialView(controller.ControllerContext, partialName);

    if (result.View == null)
    {
        throw new InvalidOperationException(
        string.Format("The partial view '{0}' could not be found", partialName));
    }
    var partialPath = ((RazorView)result.View).ViewPath;

    vp.ViewData.Model = model;

    using(StringWriter sw = new StringWriter()) {
        ViewContext viewContext = new ViewContext(controller.ControllerContext, result.View, vd, controller.TempData, sw);
        result.View.Render(viewContext, sw);

        return sw.GetStringBuilder().ToString();
    }
}

Yea, I did that and moved forward. Thanks for posting reply - maybe someone else can benefit from it :). Cheers; — effkay, Mar 01 '12 at 14:03

HttpWebRequest - ASP .NET MVC 3 Passing Session State

1 Answers1