Question

我正在开发一个多用户，多房间聊天应用程序，我的模型如下所示（为简单起见省略了App模型）：

defmodule Elemental.TxChat.User do
  use Elemental.TxChat.Web, :model

  schema "users" do
    # The rooms the user is currenly logged into
    many_to_many :rooms, Elemental.TxChat.Room, join_through: "rooms_users"
    timestamps()
  end

  def changeset(struct, params \\ %{}) do
    struct
    |> cast(params, [])
    |> validate_required([])
  end
end

和

defmodule Elemental.TxChat.Room do
  use Elemental.TxChat.Web, :model

  schema "rooms" do
    field :name, :string    
    # The user id that created this room
    field :created_by, :integer
    field :created_from_app, :integer

    many_to_many :members, Elemental.TxChat.User, join_through: "rooms_users"

    timestamps()
  end

  def changeset(struct, params \\ %{}) do
    struct
    |> cast(params, [:name, :created_from_app, :created_by])
    |> validate_required([:name, :created_by, :created_from_app])
  end
end

接下来，我从iex创建了一些房间和用户（每个房间三个）。现在我想知道：假设我希望user1属于room1和room2，而user2属于room2和room3 ......怎么做？

在我看来，虽然定义模式很好，但必须有一个像user1.rooms = [room1, room2]这样的中间步骤。所以我最终在this post上看到了一个build_assoc的例子：

Ecto.build_assoc(current_user, :post)

所以这个应用程序有用户和帖子，并试图链接他们。但我不知道它是如何实现的。数据库/ Ecto如何知道哪个用户ID与哪个帖子ID相关联？

无论如何，我尝试在我的应用iex中执行此操作：

iex(46)> Ecto.build_assoc(user1, :rooms, room1)
%Elemental.TxChat.Room{__meta__: #Ecto.Schema.Metadata<:built, "rooms">,
 created_by: 1, created_from_app: 1, id: 2,
 inserted_at: #Ecto.DateTime<2016-09-16 05:00:00>,
 members: #Ecto.Association.NotLoaded<association :members is not loaded>,
 name: "room1", updated_at: #Ecto.DateTime<2016-09-16 05:00:00>}

我认为函数调用会将user1加入模型Room，使用room1中的数据来查找目标会议室。当我在输出中看到<association :members is not loaded>时，我的心沉了下去，但我想我应该检查数据库。猜猜看，我的加入表中没有条目（rooms_users）。：（

我认为很明显，模型之间的这些联系需要以某种方式创建，但我似乎已经碰壁了。怎么做？

Answer 1

修改
附注：为users_rooms表创建UserRoom模型并创建与UserRoom.changeset / 2的关联会更容易 原始回答

我做了一个小例子项目

defmodule Playground.User do use Playground.Web, :model alias __MODULE__ schema "users" do field :title, :string many_to_many :rooms, Playground.Room, join_through: "users_rooms" timestamps() end def changeset(struct, params \\ %{}) do struct |> cast(params, [:title]) |> validate_required([:title]) end def assoc_changeset(struct, params \\ %{}) do struct |> cast(params, [:title]) |> validate_required([:title]) |> add_rooms(params, struct) end defp add_rooms(changeset, params, %User{rooms: rooms}) do case params do %{add_rooms: to_be_added} when is_list(to_be_added) -> changeset |> put_assoc(:rooms, rooms ++ to_be_added) _ -> changeset end end end

工作原理：

iex(1)> u = User.changeset(%User{}, %{title: "u"}) |> Repo.insert! iex(2)> r1 = Room.changeset(%Room{}, %{title: "r1"}) |> Repo.insert! iex(3)> r2 = Room.changeset(%Room{}, %{title: "r2”}) |> Repo.insert! iex(4)> u = User.changeset(u, %{title: "u1"}) |> Repo.update! %Playground.User{ rooms: #Ecto.Association.NotLoaded<association :rooms is not loaded>, title: "u1", ... } iex(5)> u = User.assoc_changeset(u, %{add_rooms: [r1]}) |> Repo.update! error about #Ecto.Association.NotLoaded<association :rooms is not loaded> here iex(6)> u = Repo.preload(u, :rooms) %Playground.User{ rooms: [], title: "u1", ... } iex(7)> u = User.assoc_changeset(u, %{add_rooms: [r1]}) |> Repo.update! %Playground.User{ rooms: [ %Playground.Room{title: "r1", ...} ], title: "u1", ... } iex(7)> u = User.assoc_changeset(u, %{add_rooms: [r2]}) |> Repo.update! %Playground.User{ rooms: [ %Playground.Room{title: "r1", ...}, %Playground.Room{title: "r2", ...} ], title: "u1", ... }

仍有改进的余地。帮助函数add_rooms / 3可能应该从changeset而不是第三个参数中取出用户房间并更改为add_rooms / 2。
您需要决定还需要改进哪些内容。

无法在Ecto中创建多对多关联

1 个答案: