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?
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7999 次
2 回答
17
有一个处理1-1情况的简单示例。
想象一下,我们有一个Car
和一个Engine
模型,显然还有一个Car
has_one Engine
。所以有汽车型号的代码
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
和发动机型号
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
以及一篇关于该主题的有趣博客文章,也可以澄清一些事情 ->这里
于 2015-08-17T15:48:15.033 回答
3
在一段has_many
关系中遇到同样的问题。不幸的是, aCar
不能有 many Engines
,所以我会在这篇博文中举同样的例子, of a TodoList
, with manyTodoItems
TodoList
模型:
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
模型:
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
这是表单创建 a TodoList
。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 %>
这就是它的TodoListController
样子。该create
方法是最棘手的。我不得不深入研究 Ecto Tests 以找到一种方法来完成这项工作。关联
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
于 2016-12-21T11:47:35.287 回答