DynamicSupervisors的工作人员姓名

问题描述

大家好，

我对Elixir有点陌生，并且在为Elixir中的工人设置工人名称和ID时非常迷失了，我希望有人能对我有所帮助。

我的申请文件

defmodule Squareup.Application do
  # See https://hexdocs.pm/elixir/Application.html
  # for more @R_803_4045@ion on OTP Applications
  @moduledoc false

  use Application

  def start(_type,_args) do
    children = [
      # Starts a worker by calling: Squareup.Worker.start_link(arg)
      # {Squareup.Worker,arg}
      journey,Squareup.journeySupervisor,{Squareup.DynamicjourneySupervisor,strategy: :one_for_one,name: Squareup.DynamicjourneySupervisor}
    ]

    # See https://hexdocs.pm/elixir/Supervisor.html
    # for other strategies and supported options
    opts = [strategy: :one_for_one,name: Squareup.Supervisor]
    Supervisor.start_link(children,opts)
  end
end

因此，我启动了2个主管，一个是常规主管，另一个是动态主管：

常规的：

defmodule Squareup.journeySupervisor do

  use Supervisor

  def start_link(opts) do
    Supervisor.start_link(__MODULE__,opts,name: :my_test_123)
  end

  @impl true
  def init(_init_arg) do
    children = [
      # Starts a worker by calling: Squareup.Worker.start_link(arg)
      # {Squareup.Worker,arg}
      {Queue,[:child_one]},{Queue,[:child_two]},]
    Supervisor.init(children,strategy: :one_for_one)
  end

  def child_spec(opts) do
    %{
      id: :my_test_123s,start: {__MODULE__,:start_link,[opts]},shutdown: 5_000,restart: :permanent,type: :supervisor
    }
  end
end

动态一：

defmodule Squareup.DynamicjourneySupervisor do
  use DynamicSupervisor

  def start_link(init_arg) do
    DynamicSupervisor.start_link(__MODULE__,init_arg,name: __MODULE__)
  end

  def start_child(init_args) do
    # If MyWorker is not using the new child specs,we need to pass a map:
    # spec = %{id: MyWorker,start: {MyWorker,[foo,bar,baz]}}
    spec = {Queue,init_args}
    DynamicSupervisor.start_child(__MODULE__,spec)
  end

  @impl true
  def init(init_arg) do
    DynamicSupervisor.init(strategy: :one_for_one)
  end
end

普通主管一个名为Queue

的模块

defmodule Queue do
  require Logger
  use GenServer

  ### GenServer API
  def init(state),do: {:ok,state}
  def handle_call(:dequeue,_from,[value | state]) do {:reply,value,state} end
  def handle_call(:dequeue,[]),do: {:reply,nil,[]}
  def handle_call(:queue,state),state,state}
  def handle_cast({:enqueue,value},state) do {:noreply,state ++ [value]} end

  ### Client API / Helper functions

  def start_link(state \\ []) do
    Logger.info("Initializing queue with state:")
    Logger.info(inspect(state))
    GenServer.start_link(__MODULE__,name: List.first(state))
  end

  def child_spec(opts) do
    Logger.info("Initializing child_spec:")
    Logger.info(inspect(opts))
    %{
      id: List.first(opts),type: :worker
    }
  end

  def queue,do: GenServer.call(__MODULE__,:queue)
  def enqueue(value),do: GenServer.cast(__MODULE__,{:enqueue,value})
  def dequeue,:dequeue)
end

启动应用程序时，我会这样：

12:55:45.744 [info]  Initializing child_spec:
12:55:45.749 [info]  [:child_one]
12:55:45.749 [info]  Initializing child_spec:
12:55:45.749 [info]  [:child_two]
12:55:45.749 [info]  Initializing queue with state:
12:55:45.749 [info]  [:child_one]
12:55:45.749 [info]  Initializing queue with state:
12:55:45.749 [info]  [:child_two]

iex(4)> Supervisor.which_children(:my_test_123)
[
  {:child_two,#PID<0.218.0>,:worker,[Queue]},{:child_one,#PID<0.217.0>,[Queue]}
]

但是当我告诉Dynamic Supervisor启动该过程

Squareup.DynamicjourneySupervisor.start_child([:my_dyname])
12:56:07.329 [info]  Initializing child_spec:
12:56:07.329 [info]  [:my_dyname]
12:56:07.330 [info]  Initializing queue with state:
12:56:07.330 [info]  [:my_dyname]

查找工作人员名称，然后它们总是:undefined

iex(5)> Supervisor.which_children(Squareup.DynamicjourneySupervisor)
[{:undefined,#PID<0.224.0>,[Queue]}]

工作人员的ID似乎设置正确。通过DynamicSupervisor启动进程时，是否可以设置名称？

已经非常感谢！

利昂

解决方法

似乎您只是误解了which_children/1函数的返回值。它返回一个列表，其中包含有关所有实际孩子的信息。

来自docs：

* id - it is always :undefined for dynamic supervisors

因此，您实际上是在正确命名进程。仅Queue模块的客户代码不正确。如果您要访问功能queue enqueue和dequeue，则应对其进行实际调用，以实际调用命名的GenServer。

我的意思是他们应该看起来像这样：

  def queue(name \\ __MODULE__),do: GenServer.call(name,:queue)
  def enqueue(name \\ __MODULE__,value),do: GenServer.cast(name,{:enqueue,value})
  def dequeue(name \\ __MODULE__),:dequeue)

这样GenServer.call/2将始终调用命名的GenServer。您也可以不按名称访问它们，而按其pids访问（实际上您可以根据需要从which_children/1函数动态访问它们）。

elixir