如何在 GraphQL 中返回一组对象，可能使用与返回单个对象相同的端点? [英] How to return an array of objects in GraphQL, possibly using the same endpoint as the one that returns a single object?

问题描述

我正在制作一个 GraphQL API，我可以通过它的 id 检索汽车对象，或者在没有提供参数的情况下检索所有汽车.

I am making a GraphQL API where I would be able to retrieve a car object by its id or retrieve all the cars when no parameter is provided.

使用下面的代码，我可以通过提供 id 作为参数来成功检索单个汽车对象.

Using the code below, I am successfully able to retrieve a single car object by supplying id as a parameter.

但是，在我期望对象数组的情况下，即当我根本不提供任何参数时，我在 GraphiQL 上没有得到任何结果.

schema.js

let cars = [
  { name: "Honda", id: "1" },
  { name: "Toyota", id: "2" },
  { name: "BMW", id: "3" }
];

const CarType = new GraphQLObjectType({
  name: "Car",
  fields: () => ({
    id: { type: GraphQLString },
    name: { type: GraphQLString }
  })
});

const RootQuery = new GraphQLObjectType({
  name: "RootQueryType",
  fields: {
    cars: {
      type: CarType,
      args: {
        id: { type: GraphQLString }
      },
      resolve(parent, args) {
        if (args.id) {
          console.log(cars.find(car => car.id == args.id));
          return cars.find(car => car.id == args.id);
        }
        console.log(cars);
        //***Problem Here***
        return cars;
      }
    }
  }
});

测试查询及其各自的结果:

Test queries and their respective results:

查询 1

{
  cars(id:"1"){
    name
  }
}

查询 1 响应(成功)

Query 1 Response (Success)

{
  "data": {
    "cars": {
      "name": "Honda"
    }
  }
}

查询 2

{
  cars{
    name
  }
}

查询 2 响应(失败)

Query 2 Response (Fail)

{
  "data": {
    "cars": {
      "name": null
    }
  }
}

任何帮助将不胜感激.

推荐答案

汽车和汽车列表实际上是两种不同的类型.一个字段一次不能解析为单个 Car 对象，另一次解析为 Car 对象数组.

A Car and a List of Cars are effectively two separate types. A field cannot resolve to a single Car object one time, and an array of Car object another.

您的查询为 name 返回 null，因为您告诉它 cars 字段将解析为单个对象，但它解析为数组.因此，它会在数组对象上寻找一个名为 name 的属性，但由于该属性不存在，它返回 null.

Your query is returning null for the name because you told it the cars field would resolve to a single object, but it resolved to an array instead. As a result, it's looking for a property called name on the array object and since one doesn't exist, it's returning null.

您可以通过几种不同的方式处理此问题.要将事情保持在一个查询中，您可以使用 filter 而不是 find 并将查询的类型更改为列表.

You can handle this in a couple of different ways. To keep things to one query, you can use filter instead of find and change the type of your query to a List.

cars: {
  type: new GraphQLList(CarType), // note the change here
  args: {
    id: {
      type: GraphQLString
    },
  },
  resolve: (parent, args) => {
    if (args.id) {
      return cars.filter(car => car.id === args.id);
    }
    return cars;
  }
}

或者，您可以将其拆分为两个单独的查询:

Alternatively, you could split this into two separate queries:

cars: {
  type: new GraphQLList(CarType),
  resolve: (parent, args) => cars,
},
car: {
  type: CarType,
  args: {
    id: {
      // example of using GraphQLNonNull to make the id required
      type: new GraphQLNonNull(GraphQLString)
    },
  },
  resolve: (parent, args) => cars.find(car => car.id === args.id),
}

查看文档以获取更多示例和选项.

Check the docs for more examples and options.

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