2

所以我有一个循环运行并执行某些操作的长生不老药程序。我已经声明了一个代理来创建一个 Zookeeper 连接实例,然后在循环中使用它。以下是我的代码:-

mix.exs

def application do
[ 
  applications: [:logger, :zookeeper],
  mod: {ServiceMonitor, []}
]

end    

defp deps do
[
  {:zookeeper, github: "vishnevskiy/zookeeper-elixir"}
]
end

service_monitor.ex

defmodule ServiceMonitor do
  def start(_type, _args) do
    {:ok, zk_agent} = Agent.start_link(fn -> ServiceMonitor.Registry.get_zk end)
    ServiceMonitor.Registry.start_process(zk_agent)
   end
end 

service_monitor/registry.ex

defmodule ServiceMonitor.Registry do
  alias Zookeeper.Client, as: ZK
  def start_process(zk) do

    pid = spawn_link(fn ->
        {:ok, data} = ZK.get_children(zk, "/test")
        IO.inspect(data)
    end)

    start_process(zk)

  end

  def get_zk do
    {:ok, zk} = ZK.start("localhost:2181")
    zk
  end
end

现在,如果我打印 zk 的 pid,我总是得到相同的结果,这意味着总是返回相同的 zk 实例。但我收到以下错误:-

12:44:39.647 [error] GenServer #PID<0.165.0> terminating
** (stop) bad call: #Operation performed by zk
(elixir) lib/gen_server.ex:420: Agent.Server."handle_call (overridable 1)"/3
(stdlib) gen_server.erl:629: :gen_server.try_handle_call/4
(stdlib) gen_server.erl:661: :gen_server.handle_msg/5
(stdlib) proc_lib.erl:240: :proc_lib.init_p_do_apply/3
Last message: #
State: {:ok, #PID<0.166.0>} 

另外,如果我总是在循环中初始化 zk 而不是从代理中引用,我的代码工作正常。

PS:-要重现它,您需要设置 zookeeper

4

1 回答 1

0

似乎我错误地为生成的进程运行了循环。Dogbert 关于如何在循环中获取代理返回的值的解决方案确实有效。现在,我正在生成的进程上运行循环,如下所示: -

def start_process(zk_agent) do
    spawn(ServiceMonitor.Registry, :start,[zk_agent])
    :timer.sleep(@sleep_time)
    start_process(zk_agent)
end

def start_process(zk_agent) do
    zk = Agent.get(zk_agent, &(&1))
    #Other logics
end
于 2016-09-13T05:48:35.857 回答