如何通过注册孩子获取注册的密钥（pid）

Question

我想使用Registry模块来注册动态创建的子进程。所以我将注册表添加到我的主管树中：

def init([]) do
  children = [
    supervisor(Registry, [:unique, :my_registry]),
    supervisor(MyManager, []),
    # ...
  ]
  opts = [strategy: :one_for_one, name: MySupervisor]
  supervise(children, opts)
end

孩子的注册是这样发生的：

def start_link(%{id: id}) do
  GenServer.start_link(__MODULE__, [], name: {:via, Registry, {:my_registry, id}})
end

所以，我的问题是 – 如何在我的注册表中获取所有键（子 pid）。我尝试使用Registry.keys\2，但没有成功。

def get_registry_pid do
  Supervisor.which_children(MySupervisor)
  |> Enum.filter(fn {id, _, _, _} -> id == Registry end)
  |> List.first
  |> elem(1) # get a Regisry pid from a tuple {id, pid, type, modules}
end
Registry.keys(:my_registry, get_registry_pid()) # returns []
# but 
Registry.lookup(:my_registry, id) # returns PID as expected

在此先感谢您的帮助！

Answer 1

这有点晚了，但我自己也遇到了这个问题。似乎没有官方的方法可以做到这一点，但通过挖掘Registry源代码并阅读一些内容，:erts我提出了以下解决方案：

defmodule MyApp.Sensor.Registry do
  @doc """
  Lists all the registered sensors from the registry.
  """
  @spec list_sensors!(module) :: [{String.t(), pid, any}]
  def list_sensors!(registry \\ __MODULE__) do
    try do
      # The args for `lookup_element` are taken from the private function
      # `info/1` from the `Registry` module.
      {_kind, _partitions, table, _pid_ets, _} =
        :ets.lookup_element(registry, -1, 2)

      # select_all is the result of `:ets.fun2ms(&(&1))` which works from iex
      # but not when compiled for reasons beyond my comprehension
      select_all = [{:"$1", [], [:"$1"]}]
      :ets.select(table, select_all)
    catch
      :error, :badarg ->
        raise ArgumentError, "#{inspect(registry)} is not running"
    end
  end
end