elixir - 使用持久状态更正 GenServer 实现

Question

我对一些 OTP 概念不熟悉。我有 GenServer，它将向 RabbitMQ 发布事件。此 GenServer 具有状态:amqp Chanel，它在 init() 期间启动一次，并在 cast 调用之间保持不变。

defmodule Myapp.Events.AmqpTransport do

  require Logger
  use GenServer
  use AMQP

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

  def init(_opts) do
    username = get_conf(:username)
    password = get_conf(:password)
    host = get_conf(:host)
    port = get_conf(:port)
    vhost = String.replace(get_conf(:vhost), "/", "%2f")
    exchange = get_conf(:exchange)
    {:ok, conn} = Connection.open("amqp://#{username}:#{password}@#{host}:#{port}/#{vhost}")
    {:ok, chan} = Channel.open(conn)

    {:ok, chan}
  end

  def handle_cast({:emit, event}, chan) do
    payload = :jiffy.encode(%{event: event})
    Basic.publish(
      chan,
      get_conf(:exchange),
      get_conf(:routing_key),
      payload
    )
    {:noreply, :ok, chan}
  end

  def emit(event) do
    GenServer.cast(__MODULE__, {:emit, event})
  end

  defp get_conf(key) do
    conf = Application.get_env(:myapp_events, :rabbit)
    conf[key]
  end
end

当我使用 Myapp.Events.AmqpTransport.emit(%{"hop": "hej"}) 调用它时，出现错误:

[error] Supervisor 'Elixir.Myapp.Events.Supervisor' had child
        'Elixir.Myapp.Events.AmqpTransport' started with
        'Elixir.Myapp.Events.AmqpTransport':start_link([])
        at <0.7024.0> exit with reason timeout_value
        in gen_server:loop/7 line 437
        in context child_terminated

我在这里错过了什么？

Answer 1

您应该从 GenDerver.handle_cast/2 返回二元素元组 {:noreply, chan}，而不是三元素元组 {:noreply, :ok, chan} .

转换是异步的，因此它们不返回任何东西。您的三元素元组 {:noreply, _, _} 被视为表单中的响应

{:noreply, new_state,
    timeout() | :hibernate | {:continue, term()}}

和作为第三个元素传递的state应该是超时(它不匹配:hibernate，也不匹配元组)，但它的值是无论如何都不是超时。