asp.net MVC partial view controller action

Question

I'm very new to web app development and I thought I would start with recent technology and so I'm trying to learn asp.net as-well as the MVC framework at once. This is probably a very simple question for you, MVC professionals.

My question is should a partial view have an associated action, and if so, does this action get invoked whenever a normal page uses RenderPartial() on the partial view?

Answer 1

While you can have an action that returns a partial view, you don't need an action to render a partial view. RenderPartial takes the partial view and renders it, using the given model and view data if supplied, into the current (parent) view.

You might want an action that returns a partial view if you are using AJAX to load/reload part of a page. In that case, returning the full view is not desired since you only want to reload part of the page. In this case you can have the action just return the partial view that corresponds to that section of the page.

Standard mechanism

Making use of partial view within a normal view (no action needed)

...some html...
<% Html.RenderPartial( "Partial", Model.PartialModel ); %>
...more html..

Ajax mechanism

Reloading part of a page via AJAX (note partial is rendered inline in initial page load)

...some html...
<div id="partial">
<% Html.RenderPartial( "Partial", Model.PartialModel ); %>
</div>
...more html...

<script type="text/javascript">
   $(function() {
       $('#someButton').click( function() {
           $.ajax({
              url: '/controller/action',
              data: ...some data for action...,
              dataType: 'html',
              success: function(data) {
                 $('#partial').html(data);
              },
              ...
           });
       });
   });
</script>

Controller for AJAX

public ActionResult Action(...)
{
     var model = ...

     ...

     if (Request.IsAjaxRequest())
     {
          return PartialView( "Partial", model.PartialModel );
     }
     else
     {
          return View( model );
     }
}