Use different CSS for different layouts

Question

I found some previous questions related to this, but I'm still unable to solve my problem.

I have a prediction page. Every different tartot spread demands different CSS. I tried this:

First line of the view:

<%= stylesheet_link_tag @css_to_use %>

In the controller:

def prediction
 # some other variable management
 @css_to_use = @spread.css_to_use
end

So when I hit the form on the index page to process, it comes to this prediction action. I grab the data and generate the view. And this appears:

<div id="container">
  <div id="content">
  <link href="/stylesheets/kelta_kereszt.css?1311757386" media="screen" rel="stylesheet" type="text/css" />
<div id="kirakas_doboz">
...

So the stylesheet is included in the body tag. What do I miss?

this is from the MySQL database. The cell contains utf-8 normal english characters. In this case kelta_kerest (as it is shown in the second code snipet I included). /stylesheets/kelta_kereszt.css?1311757386 Or if you think something else, than I don't know. How can I check? — Kael
– Kael, Commented Jul 27, 2011 at 9:21
Oh okay, then it's fine. You could check with a simple <%= @css_to_use %>. I thought the problem was a different one, never mind! — slhck
– slhck, Commented Jul 27, 2011 at 9:24
Actually I've figured it out, that the source of my problem was the bad use of id and class selectors to identify the success of CSS inclusion. In other hand I wanted to put the stylesheet to the place it belongs to and this is why I accepted Petr Mikitin's answer. — Kael
– Kael, Commented Jul 27, 2011 at 9:35

bender · Accepted Answer · 2011-07-27 09:36:50Z

1

Move stylesheet_link_tag into your layout file, something like this:

!!! html
%html
  %head
    = stylesheet_link_tag @css_to_use
  %body
    = yield

Or, if you want write stylesheet_link_tag directly in your view file, you can have this in your layout:

!!! html
%html
  %head
    = yield(:header)
  %body
    = yield

and then in your view

- content_for :header do
  = stylesheet_link_tag @css_to_use

edited Jul 27, 2011 at 9:36

answered Jul 27, 2011 at 9:22

bender

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2 Comments

Kael Over a year ago

Ye, of course, accept it. Actually this wont help any case. When I ask on a page where I don't set the @css_to_use variable, this goes mad on a 'can't convert nil into String'. I had to use: <% if defined?(@css_to_use)%> around the stylesheet_link_tag.

bender Over a year ago

That's right, if the variable can be blank, you should check it first. Maybe I would use a little shorter way <%= stylesheet_link_to @css_to_use unless @css_to_use.blank? %> - it seems more "ruby" :)

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