Rails - How to fix the generated URL to be a friendly URL

I have models that looks like

search.rb:
  id eg: 101
  name eg: San Francisco

cars.rb
  id
  name

The search controller redirects user to cars.

search_controller.rb
if search.search_type=='cars'
  redirect_to :controller=>'cars', :action=>'index', :id=>search
end

A query to find list of cars from San Francisco looks like:

http://localhost/cars?id=101

I overrode to_param method in search.rb like:

search.rb
def to_param
  normalized_name = name.gsub(' ', '-').gsub(/[^a-zA-Z0-9\_\-\.]/, '')
  "#{self.id}-#{normalized_name}"
end

This works to开发者_JAVA技巧 some extent: It generates URLs that look like:

http://localhost/cars?id=101-San-Francisco

However, I would like is to generate a URL that looks like

http://localhost/cars/San-Francisco

How would I go about doing that? Do I fix the routes?

Thanks for your help.

This can be tricky if you're not prepared for it. The method that controls how a model shows up in the params is pretty straightforward, as you've identified, but that's only the half of it. Later you'll have to retrieve it, so you need to set up for that:

class Search < ActiveRecord::Base
  before_save :assign_slug

  def self.from_param(id)
    self.find_by_slug(id)
  end

  def to_param
    self.slug
  end

protected
  def assign_slug
    self.slug = self.name.gsub(' ', '-').gsub(/[^\w\-\.]/, '')
  end
end

This requires adding a slug column to your Search model in order to make it possible to look up a search by it. It may be a good idea to add an index, possibly a unique one, to help retrieve these later in an efficient manner.

In your controller you don't use find but from_param instead to do the retrieval.

You'll also need to add a custom route that takes this parameter:

match '/cars/:search', :to => 'cars#index', :as => 'cars_search'

As a note you should use the route path generator methods whenever possible because multiple routes may match the same parameters. For instance:

redirect_to cars_search_path(:search => search)