I'm sorry, my English is bad. Question: I have model Pages with columns title, description etc. I can create, change, destroy these pages. I can see the contents of the link mydomain/pages/1. I need for each page has been template and route, so I can see the content on the link, for example maydomain/contacts. How to do it? Help me please.
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1You might want to have a look at the [friendly_id](https://github.com/norman/friendly_id) gem. – spickermann Dec 10 '16 at 11:15
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One way to implement your own solution is to add this to your routes file:
get '/mydomain/:slug, to: 'pages#show'
This is a pretty general matcher, so add it to the bottom of your routes so it doesn't override others.
Then your controller show action will look something like:
def show
@page = Page.find_by_slug(params[:slug])
end
This of course assumes you have a slug column on your Pages table.
n_i_c_k
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I'm assuming by "mydomain" you mean the root url of your site (e.g. myapp.example.com)
I'd suggest that you separate the problem into two parts:
- Use an attribute other than id to identify an item in the url
- Reduce the route so that the controller does not need to be specified.
For 1, have a look at this: Rails routes with :name instead of :id url parameters Note, that as @spickermann suggests friendly_id could be a good solution for you.
For 2, you will need to create a route without the controller name, and then specify the controller in the route definition. (See the Rails Routing Guide):
get ':param', to: :show, controller: 'pages'
For that to work, you will need to put it after (lower in routes.rb) so that it doesn't intefer with other routes. I'd also recommend adding a constraint to the route - to limit the wrong urls that could be routed to that rout.
