仅在加入 Phoenix 时验证某些频道

Question

如何更改此代码以连接到不需要身份验证的频道，同时仍允许在某些频道上进行身份验证？

phoenix.js:701 WebSocket 连接到 'ws://localhost:4000/socket/websocket?token=&vsn=1.0.0' 失败：WebSocket 握手期间出错：意外响应代码：403

user_socket.ex

defmodule App.UserSocket do
  use Phoenix.Socket

  ## Channels
  channel "collection:*", App.CollectionChannel

  ## Transports
  transport :websocket, Phoenix.Transports.WebSocket
  # transport :longpoll, Phoenix.Transports.LongPoll

  # Socket params are passed from the client and can
  # be used to verify and authenticate a user. After
  # verification, you can put default assigns into
  # the socket that will be set for all channels, ie
  #
  #     {:ok, assign(socket, :user_id, verified_user_id)}
  #
  # To deny connection, return `:error`.
  #
  # See `Phoenix.Token` documentation for examples in
  # performing token verification on connect.
  @max_age 2 * 7 * 24 * 60 * 60
  def connect(%{"token" => token}, socket) do
    case Phoenix.Token.verify(socket, "user socket", token, max_age: @max_age) do
      {:ok, user_id} ->
        {:ok, assign(socket, :user_id, user_id)}
      {:error, _reason} ->
        :error
    end
  end

  def connect(_params, _socket), do: :error

  # Socket id's are topics that allow you to identify all sockets for a given user:
  #
  #     def id(socket), do: "users_socket:#{socket.assigns.user_id}"
  #
  # Would allow you to broadcast a "disconnect" event and terminate
  # all active sockets and channels for a given user:
  #
  #     Style.Endpoint.broadcast("users_socket:#{user.id}", "disconnect", %{})
  #
  # Returning `nil` makes this socket anonymous.
  def id(socket), do: "users_socket:#{socket.assigns.user_id}"
end

collection_channel.ex

defmodule App.CollectionChannel do
  use App.Web, :channel

  def join("collection:lobby", _params, socket) do
    {:ok, socket}
  end
end

app.js

import socket from "./socket"

let channel = socket.channel("collection:lobby", {});

channel.join()
  .receive("ok", resp => console.log("joined the collection channel", resp))
  .receive("error", reason => console.log("join failed", reason));

socket.js

import {Socket} from "phoenix"

let socket = new Socket("/socket", {params: {token: window.userToken}})

Answer 1

要删除身份验证，而不是验证令牌，请始终在 connect 回调中返回一个 {:ok, socket} 元组：

  def connect(_params, socket) do
    {:ok, socket}
  end

并让您的id 回调返回nil，因为所有套接字都是匿名的

def id(socket), do: nil

编辑：基于通道的身份验证

如果您想同时拥有经过身份验证的频道和匿名频道，则必须在频道的 join/3 回调中处理身份验证，或者通过套接字分配指定，如果允许用户加入频道。

例如：

  def connect(%{"token" => token}, socket) do
    case Phoenix.Token.verify(socket, "user socket", token, max_age: @max_age) do
      {:ok, user_id} ->
        {:ok, assign(socket, :user_id, user_id)}
      {:error, _reason} ->
        :error
    end
  end
  def connect(_params, socket), do: {:ok, socket}

这将允许任何人加入，但仅在通过身份验证时设置user_id

# authenticated_channel.ex
defmodule App.AuthenticatedChannel  do
  use App.Web, :channel

  def join("authenticated:lobby", _params, socket) do
    if socket.assigns[:user_id] do
      {:ok, socket}
    else
      {:error, %{reason: "unauthorized"}}
  end

end

# unauthenticated_channel.ex
defmodule App.UnauthenticatedChannel do
  use App.Web, :channel

  def join("unauthenticated:lobby", _params, socket) do
    {:ok, socket}
  end
end