使用makeRemoteExecutableSchema拼接安全订阅

Question

我们已经实现了模式拼接，其中GraphQL服务器从两个远程服务器获取模式并将它们拼接在一起。当我们只使用Query和Mutations时，一切都运行正常，但现在我们有一个用例，我们甚至需要缝合订阅，远程模式已经通过它实现了auth。

我们很难搞清楚如何通过网关将connectionParams中收到的授权令牌从客户端传递到远程服务器。

这就是我们如何反省架构：

API网关代码：

const getLink = async(): Promise<ApolloLink> => {
const http = new HttpLink({uri: process.env.GRAPHQL_ENDPOINT, fetch:fetch})

const link = setContext((request, previousContext) => {
    if (previousContext
        && previousContext.graphqlContext
        && previousContext.graphqlContext.request
        && previousContext.graphqlContext.request.headers
        && previousContext.graphqlContext.request.headers.authorization) {
        const authorization = previousContext.graphqlContext.request.headers.authorization;
        return {
            headers: {
                authorization
            }
        }
    }
    else {
        return {};
    }
}).concat(http);

const wsLink: any = new WebSocketLink(new SubscriptionClient(process.env.REMOTE_GRAPHQL_WS_ENDPOINT, {
    reconnect: true,
    // There is no way to update connectionParams dynamically without resetting connection
    // connectionParams: () => { 
    //     return { Authorization: wsAuthorization }
    // }
}, ws));


// Following does not work
const wsLinkContext = setContext((request, previousContext) => {
    let authToken = previousContext.graphqlContext.connection && previousContext.graphqlContext.connection.context ? previousContext.graphqlContext.connection.context.Authorization : null
    return {
        context: {
            Authorization: authToken
        }
    }
}).concat(<any>wsLink);

const url = split(({query}) => {
    const {kind, operation} = <any>getMainDefinition(<any>query);
    return kind === 'OperationDefinition' && operation === 'subscription'
},
wsLinkContext,
link)

return url;
}

const getSchema = async (): Promise < GraphQLSchema > => {
  const link = await getLink();
  return makeRemoteExecutableSchema({
    schema: await introspectSchema(link),
    link,
  });
}
const linkSchema = `
  extend type UserPayload {
    user: User
  }
`;
const schema: any = mergeSchemas({
  schemas: [linkSchema, getSchema],
});
const server = new GraphQLServer({
  schema: schema,
  context: req => ({
    ...req,
  })
});

有没有办法用graphql-tools实现这个目标？任何帮助赞赏。

Answer 1

我有一个有效的解决方案：我的想法是不为整个应用程序创建一个SubscriptionClient实例。相反，我正在为代理服务器的每个连接创建客户端：

server.start({
    port: 4000,
    subscriptions: {
      onConnect: (connectionParams, websocket, context) => {
        return {
          subscriptionClients: {
            messageService: new SubscriptionClient(process.env.MESSAGE_SERVICE_SUBSCRIPTION_URL, {
              connectionParams,
              reconnect: true,
            }, ws)
          }
        };
      },
      onDisconnect: async (websocket, context) => {
        const params = await context.initPromise;
        const { subscriptionClients } = params;
        for (const key in subscriptionClients) {
          subscriptionClients[key].close();
        }
      }
    }
  }, (options) => console.log('Server is running on http://localhost:4000'))

如果你有更多的远程模式，你只需要在SubscriptionClient地图中创建更多的subscriptionClients实例。

要在远程模式中使用这些客户端，您需要做两件事：

1. 在上下文中公开它们：

   const server = new GraphQLServer({
     schema,
     context: ({ connection }) => {
       if (connection && connection.context) {
         return connection.context;
       }
     }
   });

2. 使用自定义链接实现而不是WsLink

   (operation, forward) => {
       const context = operation.getContext();
       const { graphqlContext: { subscriptionClients } } = context;
       return subscriptionClients && subscriptionClients[clientName] && subscriptionClients[clientName].request(operation);
   };

这样，整个连接参数将被传递给远程服务器。

整个例子可以在这里找到：https://gist.github.com/josephktcheung/cd1b65b321736a520ae9d822ae5a951b

免责声明：

代码不是我的，因为@josephktcheung提供了一个例子。我只是帮了一点。这是最初的讨论：https://github.com/apollographql/graphql-tools/issues/864