为什么我没有为#Ecto.Query实现协议可枚举？

Question

虽然我将查询传递给存储库灵感来自这样的答案:

teams_users = Repo.all (from(t in Team, where: t.owner_id == ^user_id))
    |> Enum.each( &team_users/1 )


def team_users (team) do
    %{id: id} = team
    Repo.all (from(tu in TeamUser, where: tu.team_id == ^id))
end

但是,我收到了这个错误:

[error] GenServer #PID<0.450.0> terminating
** (Protocol.UndefinedError) protocol Enumerable not implemented for #Ecto.Query<from t in App.Team, where: t.owner_id == ^1>
    (elixir) lib/enum.ex:1: Enumerable.impl_for!/1
    (elixir) lib/enum.ex:116: Enumerable.reduce/3
    (elixir) lib/enum.ex:1477: Enum.reduce/3
    (elixir) lib/enum.ex:609: Enum.each/2
    (App) web/channels/user_channel.ex:93: App.UserChannel.handle_in/3
    (phoenix) lib/phoenix/channel/server.ex:223: Phoenix.Channel.Server.handle_info/2
    (stdlib) gen_server.erl:615: :gen_server.try_dispatch/4
    (stdlib) gen_server.erl:681: :gen_server.handle_msg/5
    (stdlib) proc_lib.erl:240: :proc_lib.init_p_do_apply/3

我想要做的是获得所有团队,然后获得每个团队的用户,然后我希望将所有用户放在一个阵列中.

我错过了什么吗？任何建议？有没有更好的方法来实现这一目标？

Answer 1

所以这里有一个问题:

teams_users = Repo.all (from(t in Team, where: t.owner_id == ^user_id))
|> Enum.each( &team_users/1 )

您应该删除的空间之间Repo.all和(

teams_users = Repo.all(from(t in Team, where: t.owner_id == ^user_id))
|> Enum.each( &team_users/1 )

你也可以这样写:

teams_users =
from(t in Team, where: t.owner_id == ^user_id)
|> Repo.all()
|> Enum.each( &team_users/1 )

但是,这样做会引入n + 1查询.您将进行一次查询以获取您的团队,然后另一次查询以获取团队用户.您应该查看Repo.preload/2.

teams_users =
  from(t in Team, where: t.owner_id == ^user_id)
  |> Repo.all()
  |> Repo.preload(:team_users)