如何在Phoenix框架中处理关联和嵌套表单？

Question

在Phoenix框架中处理关联和嵌套表单的方法是什么？如何创建具有嵌套属性的表单？如何在控制器和模型中处理它？

Answer 1

这里有一个处理1-1情况的简单示例。

假设我们有一个Car和一个Engine模型，显然还有一个Car has_one Engine。所以这里有汽车模型的代码

defmodule MyApp.Car do
  use MyApp.Web, :model

  schema "cars" do
    field :name, :string            

    has_one :engine, MyApp.Engine

    timestamps
  end

  def changeset(model, params \\ :empty) do
    model
    |> cast(params, ~w(name), ~w())
    |> validate_length(:name, min: 5, message: "No way it's that short")    
  end

end

和引擎模型

defmodule MyApp.Engine do
  use MyApp.Web, :model

  schema "engines" do
    field :type, :string            

    belongs_to :car, MyApp.Car

    timestamps
  end

  def changeset(model, params \\ :empty) do
    model
    |> cast(params, ~w(type), ~w())
    |> validate_length(:type, max: 10, message: "No way it's that long")    
  end

end

表单->的简单模板

<%= form_for @changeset, cars_path(@conn, :create), fn c -> %>

  <%= text_input c, :name %>

  <%= inputs_for c, :engine, fn e -> %>

    <%= text_input e, :type %>

  <% end %>  

  <button name="button" type="submit">Create</button>

<% end %>

和控制器->

defmodule MyApp.CarController do
  use MyApp.Web, :controller
  alias MyApp.Car
  alias MyApp.Engine

  plug :scrub_params, "car" when action in [:create]

  def new(conn, _params) do    
    changeset = Car.changeset(%Car{engine: %Engine{}})    
    render conn, "new.html", changeset: changeset
  end

  def create(conn, %{"car" => car_params}) do    
    engine_changeset = Engine.changeset(%Engine{}, car_params["engine"])
    car_changeset = Car.changeset(%Car{engine: engine_changeset}, car_params)
    if car_changeset.valid? do
      Repo.transaction fn ->
        car = Repo.insert!(car_changeset)
        engine = Ecto.Model.build(car, :engine)
        Repo.insert!(engine)
      end
      redirect conn, to: main_page_path(conn, :index)
    else
      render conn, "new.html", changeset: car_changeset
    end
  end    

end

和一篇关于这个主题的有趣的博客文章，它也可以澄清一些事情-> here

Answer 2

在has_many关系中遇到了同样的问题。不幸的是，一个Car不能有多个Engines，所以我在这个blogpost中举了同样的例子，一个TodoList有很多个TodoItems

TodoList型号：

defmodule MyApp.TodoList do
  use MyApp.Web, :model

  schema "todo_lists" do
    field :title, :string            

    has_many :todo_items, MyApp.TodoItem

    timestamps
  end
 
  def changeset(model, params \\ :{}) do
    model
    |> cast(params, [:title])
    |> cast_assoc(:todo_items)
  end 
end

TodoItem型号：

defmodule MyApp.TodoItem do
  use MyApp.Web, :model

  schema "todo_items" do
    field :body, :string

    belongs_to :todo_list, MyApp.TodoList

    timestamps
  end

  def changeset(model, params \\ :{}) do
    model
    |> cast(params, [:body])
  end
end

下面是表单创建TodoList的过程。为了简单起见，我们现在只添加一项。

<%= form_for @changeset, todo_lists_path(@conn, :create), fn f -> %>    
  <%= text_input f, :title %>  
  <%= inputs_for f, :todo_items, fn i -> %> 
    <%= text_input i, :body %> 
  <% end %>   
  <button name="button" type="submit">Create</button> 
<% end %>

这就是TodoListController的样子。create方法是最难处理的。我不得不深入研究Ecto测试，以找到一种方法来实现这一点。Link

defmodule MyApp.TodoListController do
  use MyApp.Web, :controller

  alias MyApp.TodoList
  alias MyApp.TodoItem

  def new(conn, _params) do
    todo_item = TodoItem.changeset(%TodoItem{})
    changeset = TodoList.changeset(%TodoList{todo_items: [todo_item]})

    render conn, "new.html", changeset: changeset
  end

  def create(conn, %{"todo_list" => todo_list_params}) do
    todo_item_changeset =
      TodoItem.changeset(%TodoItem{}, todo_item["todo_items"]["0"])
    changeset =
      TodoList.changeset(%TodoList{}, %{title: todo_list_params["title"]})
      |> Ecto.Changeset.put_assoc(:todo_items, [todo_item_changeset])

    case Repo.insert(changeset) do
      {:ok, company} ->
        conn
        |> put_flash(:info, "TodoList created!")
        |> redirect(to: page_path(conn, :index))
      {:error, changeset} ->
        conn
        |> render "new.html", changeset: changeset
    end
  end
end