8

我想联合服务，但为了简单起见，让联合网关也拥有自己的模式和逻辑，它们将代理 REST API 端点。现在看起来我需要分别拥有联合网关服务、联合 graphql 服务和其余<->graphql 桥接服务。无论如何，在我们的例子中，rest-graphql 网关至少暂时可以存在于联邦网关中，以避免不必要的引导和维护。

看起来阿波罗联邦网关localServiceList似乎正是为了这个目的。一个示例配置：

const gateway = new ApolloGateway({
    serviceList: [
        { name: "some-service", url: "http://localhost:40001/graph" }
    ],
    localServiceList: [
        { name: "rest-bridge", typeDefs }
    ]
});

但它不能解决问题：如果有 localServiceList，它会跳过 serviceList。

所以问题是：这是否可能在 Apollo 联邦网关中也拥有自己的模式和逻辑？

4

1 回答 1

5

是的，可以这样做：

import { buildFederatedSchema } from '@apollo/federation';
import {
  ApolloGateway,
  LocalGraphQLDataSource,
  RemoteGraphQLDataSource
} from '@apollo/gateway';
import gql from 'graphql-tag';

const localServices = {
  foo: {
    schema: {
      typeDefs: gql`
        // ...
      `,
      resolvers: {
        // ...
      }
    }
  },
  bar: {
    schema: {
      typeDefs: gql`
        // ...
      `,
      resolvers: {
        // ...
      }
    }
  }
};

const remoteServices = {
  baz: {
    url: 'http://baz.local/graphql'
  },
  qux: {
    url: 'http://qux.local/graphql'
  }
};

const services = {
  ...localServices,
  ...remoteServices
};

// By providing a protocol we trick ApolloGateway into thinking that this is a valid URL;
// otherwise it assumes it's a relative URL, and complains.
const DUMMY_SERVICE_URL = 'https://';

const gateway = new ApolloGateway({
  // We can't use localServiceList and serviceList at the same time,
  // so we pretend the local services are remote, but point the ApolloGateway
  // at LocalGraphQLDataSources instead...
  serviceList: Object.keys(services).map(name => ({
    name,
    url: services[name].url || DUMMY_SERVICE_URL
  })),
  buildService({ name, url }) {
    if (url === DUMMY_SERVICE_URL) {
      return new LocalGraphQLDataSource(
        buildFederatedSchema(
          services[name].schema
        )
      );
    } else {
      return new RemoteGraphQLDataSource({
        url
      });
    }
  }
});

const apolloServer = new ApolloServer({
  gateway,
  subscriptions: false
});
于 2020-05-06T13:58:42.307 回答