Apollo Federation Gateway:在构建超图时包含本地模式

Question

在为Apollo Federation的网关构建supergraph时，您将创建一个.yaml配置文件，其中包含指向子图的路由urls。例如：https://github.com/apollographql/supergraph-demo/blob/main/subgraphs/inventory/inventory.js

//supergraph.yaml
subgraphs:
  inventory:
    routing_url: http://inventory:4000/graphql
    schema:
      file: ./subgraphs/inventory/inventory.graphql
  products:
    routing_url: http://products:4000/graphql
    schema:
      file: ./subgraphs/products/products.graphql
  users:
    routing_url: http://users:4000/graphql
    schema:
      file: ./subgraphs/users/users.graphql

在上面的例子中，他们为每个子图启动了一个Apollo服务器，并组成了一个超图。有没有可能在不启动Apollo服务器和只包含本地模式的情况下构建超图？

Answer 1

你可以的。遵循本教程：https://www.apollographql.com/blog/backend/using-apollo-federation-with-local-schemas/

不使用超图和子图，而是使用serviceList并有条件地构建数据源。

const gateway = new ApolloGateway({
  serviceList: [
    { name: "products", url: "http://localhost:4002" },
    { name: "countries", url: "http://countries" },
  ],
  buildService: ({ url }) => {
    if (url === "http://countries") {
      return new LocalGraphQLDataSource(getCountriesSchema());
    } else {
      return new RemoteGraphQLDataSource({
        url,
      });
    }
  },
});