如何在GraphQL中返回对象数组，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GraphQL API，在其中我可以通过其id检索汽车对象或在不提供任何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，在我希望有一组对象的情况下，即当我完全不提供任何参数时，在GraphiQL上将不会得到任何结果.

schema.js

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;
      }
    }
  }
});

测试查询及其相应结果:

Test queries and their respective results:

查询1

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

查询1响应(成功)

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

查询2

{
  cars{
    name
  }
}

查询2响应(失败)

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

任何帮助将不胜感激.

Any help would be much appreciated.

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一个Car对象，而另一个不能解析为一个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.

您的查询返回的name为空，因为您告诉它cars字段可以解析为单个对象，但是却可以解析为数组.结果，它将在数组对象上寻找一个名为name的属性，由于不存在该属性，因此它将返回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.

您可以通过两种不同的方式来处理此问题.要将内容保留为一个查询，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),
}

查看文档以获得更多示例和选项.

Check the docs for more examples and options.

这篇关于如何在GraphQL中返回对象数组，可能使用与返回单个对象相同的端点?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！