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 =＆amp; vsn = 1.0.0'失败：WebSocket握手期间出错：意外响应代码：403 < /强>

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{:ok, socket}回调中返回connect元组：

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

并使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：基于频道的身份验证

如果您想同时拥有身份验证通道和匿名通道，则必须在频道的join/3回调中处理身份验证，或者如果允许用户加入频道，则通过套接字指定

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，但仅在经过身份验证时设置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

仅在凤凰城加入时验证一些频道

1 个答案:

编辑：基于频道的身份验证