Question

我使用GenServer作为排队系统。如何让整个应用访问相同的流程？

我认为把它放在application.ex会这样做，就像这样：

children = [
  supervisor(Prefect, [], name: PrefectQueue)
]

我的Prefect模块是GenServer：

defmodule Prefect do
  use GenServer
  alias Prefect.Document

  # client
  def start_link() do
    GenServer.start_link(__MODULE__, [])
  end

  def add(pid, item) do
    GenServer.cast(pid, item)
  end

  # server
  def handle_cast(item, list) do
    updated_list = [item | list]
    {:noreply, updated_list}
  end
end

我似乎无法在我的控制器中访问它，但是：

defmodule PrefectWeb.DocumentController do
  use PrefectWeb, :controller

  def create(conn, params) do
    Prefect.add(PrefectQueue, params["id"])
    conn
    |> send_resp(200, "Queued #{Prefect.view(PrefectQueue)}")
  end
end

发布到该创建函数会出现此错误：

[info] POST /api/documents
[debug] Processing with PrefectWeb.DocumentController.create/2
  Parameters: %{"id" => "1"}
  Pipelines: [:api]
[error] #PID<0.349.0> running PrefectWeb.Endpoint terminated
Server: 0.0.0.0:4000 (http)
Request: POST /api/documents
** (exit) exited in: GenServer.call(Prefect.Queue, :view, 5000)
    ** (EXIT) no process: the process is not alive or there's no process currently associated with the given name, possibly because its application isn't started

Answer 1

您必须在GenServer本身中命名该进程，而不是通过supervisor子列表。尝试：

defmodule Prefect do
  use GenServer
  alias Prefect.Document

  # client
  def start_link() do
    GenServer.start_link(__MODULE__, [], name: __MODULE__)
  end

  def add(pid, item) do
    GenServer.cast(pid, item)
  end

  # server
  def handle_cast(item, list) do
    updated_list = [item | list]
    {:noreply, updated_list}
  end
end

现在应该像Prefect.view(Prefect)一样调用您的流程。

如何为我的整个应用程序提供GenServer进程？

1 个答案: