如何为我的整个应用程序提供GenServer进程？

Question

我正在使用GenServer作为排队系统。我怎样才能使整个应用程序访问相同的过程？

我想过把它application.ex将做到这一点，就像这样：

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

我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 

我似乎无法访问我的控制器，虽然：

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 

发布到该创建函数给出此错误：

[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本身命名进程，而不是通过supervisor子级列表。请尝试：

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 

您的过程现在应该像这样调用Prefect.view(Prefect)。