How to define a route like /properties/filter_objects alongside /properties/:id

Questions & Help Questions
, ,
1

I want to have two GET routes working like this:

a. /properties/:id
b. /properties/filter_objects

a. handled by the normal show action and b. by a custom controller action. I have achieved this by doing:

  def show(conn, %{"id" => id}, user) when id != "filter_objects" do
    property = Properties.get_property!(id)
    render(conn, "show.json", property: property)
  end

  def show(conn, %{"id" => id}, user) when id == "filter_objects" do
    users = User |> Repo.all
    pretty_json(conn, %{users: users})
  end

Is there a better way to achieve this? Thank you.

2 
def show(conn, %{"id" => "filter_objects"}, user) do
    users = User |> Repo.all
    pretty_json(conn, %{users: users})
  end

  def show(conn, %{"id" => id}, user) do
    property = Properties.get_property!(id)
    render(conn, "show.json", property: property)
  end
3

@LostKobrakai

Thank you. This seems fine. I thought the Phoenix way to do this is by some definitions in router.ex. Your suggestion makes things simpler.

One question, is the order of function definitions important?

4

The phoenix way to do this ‘is’ in the Router. ^.^

Put the specific route /properties/filter_objects first then the generic route /properties/:id second so it gets matched on in order, can even have them go to different functions. :slight_smile:

5

I have managed to achieve that by:

  scope "/api", RentalWeb.Api do
    pipe_through :api
    
    get "/properties/filter_objects", PropertyController, :filter_objects
    resources "/properties", PropertyController
  end``