Rails: Render a View (not a partial) From Within a View

nullnullnull picture nullnullnull · May 2, 2013 · Viewed 33.5k times · Source

I have a controller that responds to both html and js. The html view renders the whole page (including the header and footer), while the js only replaces #main. Aside from the header and footer, both formats render the same content. I can get this effect with three files:

_show.html.erb
<div>Content!</div>

show.html.erb
<%= render "show" %>

show.js.erb
$("#main").fadeIn("<%= escape_javascript(render 'show') %>");

This works, but I'd prefer if I didn't need a separate _show partial. Unfortunately, this doesn't work:

show.html.erb
<div>Content!</div>

show.js.erb
$("#main").fadeIn("<%= escape_javascript(render 'show') %>");

As Rails will look for the show partial, not the actual view.

Is there a way to get Rails to look for the view file, rather than a partial?

Answer

numbers1311407 picture numbers1311407 · May 2, 2013

Rendering a non-partial view inside another view isn't exactly the Rails Way™. Your current solution is probably better, and not an uncommon approach. Rename it _body, or something else appropriate, if you feel weird about the the partial name being the same as the action.

However if your view can be shared, as it seems like it could in this case, you could just make it a layout.

This is facilitated by the fact that, somewhat against the principle of least surprise, Rails will render an html template for a js action if no js template exists. This means that you could remove both the js template, and the partial, and just create a layout entitled, for example, fadein.js.erb:

# yourviews/show.html.erb
<div>Content!</div>

# layouts/fadein.js.erb
$("#main").fadeIn("<%= escape_javascript(yield) %>");

# YourController.rb
def show
  # ...
  respond_to do |wants|
    wants.html
    wants.js { render :layout => "fadein" }
  end
end