Render action return View(); form problem - asp.net-mvc

Question

I'm new to MVC, so please bear with me. :-)
I've got a strongly typed "Story" View. This View (story) can have Comments.
I've created two Views (not partials) for my Comments controller "ListStoryComments" and "CreateStoryComment", which do what their names imply. These Views are included in the Story View using RenderAction, e.g.:
<!-- List comments -->
<h2>All Comments</h2>
<% Html.RenderAction("ListStoryComments", "Comments", new { id = Model.Story.Id }); %>
<!-- Create new comment -->
<% Html.RenderAction("CreateStoryComment", "Comments", new { id = Model.Story.Id }); %>
(I pass in the Story id in order to list related comments).
All works as I hoped, except, when I post a new comment using the form, it returns the current (parent) View, but the Comments form field is still showing the last content I typed in and the ListStoryComments View isn’t updated to show the new story.
Basically, the page is being loaded from cache, as if I had pressed the browser’s back button. If I press f5 it will try to repost the form. If I reload the page manually (reenter the URL in the browser's address bar), and then press f5, I will see my new content and the empty form field, which is my desired result.
For completeness, my CreateStoryComment action looks like this:
[HttpPost]
public ActionResult CreateStoryComment([Bind(Exclude = "Id, Timestamp, ByUserId, ForUserId")]Comment commentToCreate)
{
try
{
commentToCreate.ByUserId = userGuid;
commentToCreate.ForUserId = userGuid;
commentToCreate.StoryId = 2; // hard-coded for testing
_repository.CreateComment(commentToCreate);
return View();
}
catch
{
return View();
}
}

Answer 1

The answer is to use return RedirectToAction(). Using this enforces the PRG pattern and achieves my goal.
My earlier comment to my original post did cause an error, that I guess I'm still confused about, but this works:
return RedirectToAction("Details", "Steps", new { id = "2" });

Answer 2

I, and this is a personal opinion, think you've tackled this the wrong way.
Personally I would;
Create a view called ListStories.
Create a partial view that lists the
stories.
Create a partial view to create a
story.
When you want to add a story, simply
show the add story html.
Then when the user presses a button
you do a jQuery postback, add the
new story and return a PartialView
of either the new story or all the
stories.
If you return a partial view of all
stories then replace the bounding
div that contains all the stories
with the new data.
If you return only a single story
then append it to the end of the div
containing the stories.
I know this means a lot of re-work and it sounds complex and like a lot of work but doing it like this means greater flexibility later on because you can re-use the partial views or you can make a change once and all views using that partial view are now updated.
also, using jQuery means that the adding of stories looks seemless w/out any obvious post back which is nice.

Answer 3

Since the problem seems to be caching, you can simply disable/limit caching.
Add the following attribute to your actions:
[OutputCache(Duration = 0, VaryByParam = "none")]
This will tell the browser to cache the page, but for 0 seconds. When the post reloads the page, you should see the desired results.

Answer 4

The answer is to make sure your form action is properly set. If you have used renderaction and not set the form controller and action manually then the action will be the current URL.
Try:
<% using (Html.BeginForm("ActionName", "ControllerName")) {%>
Instead of:
<% using (Html.BeginForm()) {%>