如何使用自定义返回类型在 type-graphql 中创建 GraphQL 查询答案

【问题标题】：How to create a GraphQL query in type-graphql with a custom return type如何使用自定义返回类型在 type-graphql 中创建 GraphQL 查询
【发布时间】：2020-12-19 18:28:42
【问题描述】：

我正在尝试创建一个 graphql 查询来查询两个数据库表并返回一个自定义对象。我很难弄清楚如何定义这种自定义类型。

我在 typeorm 中创建了两个实体：Track 和 Vehicle：

import { Entity, PrimaryGeneratedColumn, Column, BaseEntity } from "typeorm";
import { ObjectType, Field, ID } from "type-graphql";

@ObjectType()
@Entity()
export class Track extends BaseEntity {
  @Field(() => ID)
  @PrimaryGeneratedColumn()
  id: number;

  @Field()
  @Column("text", { unique: true })
  trackId: string;

  @Field()
  @Column()
  origin: string;

  @Field()
  @Column("text", { unique: true })
  name: string;
}

import { Entity, PrimaryGeneratedColumn, Column, BaseEntity } from "typeorm";
import { ObjectType, Field, ID } from "type-graphql";

@ObjectType()
@Entity()
export class Vehicle extends BaseEntity {
  @Field(() => ID)
  @PrimaryGeneratedColumn()
  id: number;

  @Field()
  @Column("text", { unique: true })
  vehicleId: string;

  @Field()
  @Column()
  vehicleFolder: string;

  @Field()
  @Column("text", { unique: true })
  name: string;
}

我想做一个返回如下对象的查询：

interface TracksAndVehicles {
  tracks: Track[];
  vehicles: Vehicle[];

我不知道该怎么做。这是我的解析器：

import { Resolver, Query, Field, ObjectType } from "type-graphql";
import { Track } from "../../entity/Track";
import { Vehicle } from "../../entity/Vehicle";

@ObjectType()
class TracksAndVehicles {
  @Field()
  tracks: Track[];

  @Field()
  vehicles: Vehicle[];
}

@Resolver()
export class GetTracksAndVehiclesResolver {
  @Query(() => TracksAndVehicles)
  async getTracksAndVehicles(): Promise<TracksAndVehicles> {
    const tracks = await Track.find();
    const vehicles = await Vehicle.find();
    return { tracks, vehicles };
  }
}

我收到此错误：Error: You need to provide explicit type for TracksAndVehicles#tracks

有什么想法吗？

【问题讨论】：

标签： typescript graphql typeorm typegraphql

【解决方案1】：

https://typegraphql.com/docs/types-and-fields.html

  @Field(type => [Track])
  tracks: Track[];

【讨论】：