更改GraphQL响应的结构

问题描述

我有一个使用以下现有路线的GraphQL和Express项目:

I have a GraphQL and Express project with these existing routes:

// ...

const graphiqlExpress = require('graphql-server-express').graphiqlExpress;
const graphqlExpress = require('graphql-server-express').graphqlExpress;
const makeExecutableSchema = require('graphql-tools').makeExecutableSchema;

// ...

const schema = makeExecutableSchema({
  typeDefs,
  resolvers,
});

// ...

app.use('/graphql', bodyParser.json(), graphqlExpress({ schema }));
app.use(
  '/graphiql',
  graphiqlExpress({
    endpointURL: '/graphql',
  })
);

在 http://localhost:3030/graphql?query = {regions(countries:["FR"]，level:0){...} 上获得GET后的结果正常的GraphQL响应:

My results after a GET on http://localhost:3030/graphql?query={regions(countries:["FR"],level:0){...} look like a normal GraphQL response:

{
  "data": {
    "regions": [
      {
        "type": "FeatureCollection",
        "features": [
          ...
          ...
        ]
      }
    ]
  }
}

是否可以将响应转换为更有效的GeoJSON格式?(没有"数据:{} "，并且没有我的查询名称等)，例如:

Is there a way to transform the response to something more like a valid GeoJSON format? (no "data:{ }" and without the name of my query, etc.) such as:

{
  "type": "FeatureCollection",
    "features": [
      ...
      ...
    ]  
}

我已经想到要做的是使用 middleare (但是如何?)和/或诸如 graphql-normalizr 之类的规范化器(但我不这样做)不知道如何用Express插入)

What I've already thought of doing is either using a middleare (but how?) and/or a normalizer such as graphql-normalizr (but I don't know how to plug it with Express)

推荐答案

经过左右几个小时的搜索，我找到了可行的解决方案.幸运的是，您可以使用 formatResponse : https://github.com/apollographql/apollo-server/blob/master/packages/apollo-server-core/src/graphqlOptions.ts#L43

After a few good hours of searching left and right I found a viable solution.Luckily there is a formatResponse that you can use: https://github.com/apollographql/apollo-server/blob/master/packages/apollo-server-core/src/graphqlOptions.ts#L43

更新后的代码如下:

app.use(
  '/graphql',
  bodyParser.json(),
  graphqlExpress({ schema, formatResponse: res => (res.data.regions && res.data.regions[0]) || res })
);

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