为什么我的 graphql 嵌套查询返回 null?

Why is my graphql nested query returning null?

我是 graphql 的新手 运行 遇到嵌套查询问题,需要帮助传递 id 以识别关系。

查询

您可以在上方看到 PERFORMED_FOR_Affiliation 为空,尽管它在架构中定义为 Affiliation 类型。

type Query {
      affiliations(affn_id: ID!): [Affiliation]
      performances(pfrm_id: ID!): [Performance]
      PERFORMED_FOR_Affiliation(affn_id: ID!): Affiliation
      Performance_PERFORMED_FOR(pfrm_id: ID!): [Performance]
    }

PERFORMED_FOR_Affiliation 查询类似于 affiliations 查询,仅关系应该 return 只有 1 个隶属关系(具有匹配的 uid)。

我假设 affn_id 没有被正确传递并且不确定如何正确地传递。 PERFORMED_FOR_Affiliation 是否需要自己的模式?

架构

type Performance {
    pfrm_id: ID!
    mark: Int
    affn_id: ID!
    PERFORMED_FOR_Affiliation: Affiliation
}

type Affiliation {
    affn_id: ID!
    name: String
    Performance_PERFORMED_FOR: [Performance]
}

我见过一些使用 'nodes' 和 'edge' 类型的模式。由于我有许多其他关系,这会是定义图表的更好方法吗?

解析器

import performances from './mockData/performances.js';
import affiliations from './mockData/affiliations.js';

export const resolvers = {
  Query: {
    affiliations: (root, args) => {
      return affiliations;
    },
    performances: (root, args) => {
      return performances;
    },
    PERFORMED_FOR_Affiliation: (root, args) => {
      return affiliations;
    },
    Performance_PERFORMED_FOR: (root, args) => {
      return performances;
    },
  },
};

模拟数据

//affiliations.js
module.exports = [
  {
    "affn_id": "43700F3BE17145399924AC176EACBEF4",
    "name": "Richmond Senior"
  },
  {
    "affn_id": "8BDE709AC757416082950B1BEED0CE0A",
    "name": "Cedar City"
  },
  {
    "affn_id": "123D201BB17545E3B6ECCCCB5FC61FA3",
    "name": "Delta"
  }
]

// performances.js
module.exports = [
  {
    pfrm_id: "6BD41C6B1C4B43D199DE42A4A408DF1A",
    mark: 1270000,
    affn_id: "43700F3BE17145399924AC176EACBEF4",
  },
  {
    pfrm_id: "EA2FBC6AB891460EA557F5B60984AD8A",
    mark: 1422400,
    affn_id: "8BDE709AC757416082950B1BEED0CE0A",
  },
  {
    pfrm_id: "54A6EEB9552C49AC9F7A87E68AC272A2",
    mark: 1422400,
    affn_id: "123D201BB17545E3B6ECCCCB5FC61FA3",
  },
]

是的,您应该在解析器中实现它。我猜你会在后台有一个数据库,例如mongoose:

您查询表演,其中您 populate 从属关系

performances: (root, args) => {
  return new Promise((resolve, reject) => {
    performanceModel.find() // find all performances
      .populate("PERFORMED_FOR_Affiliation") // add values for affiliates
      .exec((error, performances) => {
        if (!error) {
          resolve(performances); // resolve correct object
        } else {
          reject(error);
        }
      });
  });
},