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  • Aron 7 posts 107 karma points
    Jun 15, 2016 @ 11:24
    Aron
    0

    Creating a search-form in umbraco using a surface controller

    Using Umbraco And Getting Started
    Umbraco 7

    Hello!

    I'm currently trying to implement a simple search page in umbraco using a surface controller. What I've got so far is as follows:

    • View (created from the umbraco backoffice with a doc type and template)

      • The view calls Html.Action("GetSearchForm", "MyController") which in turn returns a partialview (_SearchForm) containing the form.

      • _SearchForm should render _SearchResults, that binds to _SearchForms Model.Results @Html.Partial("_SearchResults, Model.Results

      • I'm currently using Html.BeginUmbracoForm("FormSubmit"...) with FormMethod set to post.

    Posting to the action works fine, but when I've retrieved the results from the database (not the umbraco database) I want to return the partialview updated with the results, my controller looks something like this:

        [HttpPost]
    public ActionResult FormSubmit(SearchModel model)
    {
          if(!ModelState.IsValid)
          {
              return CurrentUmbracoPage();
          }

          //Results is a List of type Result
          model.Results =  GetResults();

          return PartialView("_SearchForm", model);
    }

    When I do this I end up losing the layout of the page, Sure, i get the partial-view rendered, and the results are there, but it's obviously wrong.

    What am i doing wrong? As of now, i've resorted to putting the results into

    TempData["Results"] = GetResults(); return CurrentUmbracoPage();

    but it's a solution that i find repulsive.

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  • Jinesh Kaneriya 22 posts 155 karma points
    Jun 15, 2016 @ 13:19
    Jinesh Kaneriya
    0

    Use ajax call to controller via javascript or jquery like

    $.ajax({
                url: "/umbraco/surface/contact/submit",
                type: 'post',
                dataType: 'json',
                data: contactdata,
                async: true,
                success: function (response) {                  
                },
                error: function (xhr, status, error) {
                    alert(error);
                }
            });

    In ajax's success function do DOM manipulation and add search content.

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  • Aron 7 posts 107 karma points
    Jun 15, 2016 @ 13:31
    Aron
    100

    That is true, however i think i managed to fix it.

    My main view now has the following:

    ViewCreatedInUmbracoBackOffice.cshtml = 

    @inherits UmbracoViewPage<MyModel>
@{
     Layout = "Layout.cshtml";
     var currentPage = Umbraco.Content(umbraco.NodeFactory.Node.GetCurrent().Id);
 }

 @Html.Partial("_SearchForm", model)
 @Html.Partial("_SearchResults", model.Results)

    I've also set up a controller that inherits from RenderMvcController, and i've overridden the index

    //Returning the view does not trigger this as far as i can see. It only fires on the first request of the page, which is why i'd say its safe to return a new MyModel()
public override ActionResult Index(RenderModel model)
{
    //Please note that MyModel must inherit from RenderModel.
    return base.Index(new MyModel());
}

    ! Please read the following

    The controller inheriting from RenderMvcController should be named exactly as the view that was created in the backoffice

    I've created a separate controller that inherits from SurfaceController, this now handles the httppost.

    In my FormSubmit i can now do as follows:

    [HttpPost]
public ActionResult FormSubmit(SearchModel model)
{
      if(!ModelState.IsValid)
      {
          return CurrentUmbracoPage();
      }

      //Results is a List of type Result
      model.Results =  GetResults();

      return View("ViewCreatedInUmbracoBackOffice", model):
}

    All of this allows me to still use the CurrentPage.GetGridHtml(), and the Umbraco.GetDictionaryValue(), which is super handy.

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