GraphQL：如何将参数传递给子对象

Question

我正在使用 GraphQL 查询一个对象，该对象将由大约 15 个不同的 REST 调用组成。这是我从查询中传入 ID 的根查询。这适用于正确解析的主要学生对象。但是，我需要弄清楚如何将 ID 传递给地址解析器。我尝试将 args 添加到 address 对象，但出现错误，表明 args 不是从 Student 对象传递下来的。所以我的问题是：如何将参数从客户端查询传递到 GraphQL 服务器中的子对象？

let rootQuery = new GraphQLObjectType({
    name: 'Query',
    description: `The root query`,
    fields: () => ({
        Student : {
            type: Student ,
            args: {
                id: {
                    name: 'id',
                    type: new GraphQLNonNull(GraphQLString)
                }
            },
            resolve: (obj, args, ast) => {
                return Resolver(args.id).Student();
            }
        }
    })
});

export default rootQuery;

这是我的小学生对象，我link其他对象。在这种情况下，我附加了 ADDRESS 对象。

import {
GraphQLInt,
GraphQLObjectType,
GraphQLString,
GraphQLNonNull,
GraphQLList
} from 'graphql';

import Resolver from '../../resolver.js'
import iAddressType from './address.js'

let Student = new GraphQLObjectType({
    name: 'STUDENT',
    fields: () => ({
        SCHOOLCODE: { type: GraphQLString },
        LASTNAME: { type: GraphQLString },
        ACCOUNTID: { type: GraphQLInt },
        ALIENIDNUMBER: { type: GraphQLInt },
        MIDDLEINITIAL: { type: GraphQLString },
        DATELASTCHANGED: { type: GraphQLString },
        ENROLLDATE: { type: GraphQLString },
        FIRSTNAME: { type: GraphQLString },
        DRIVERSLICENSESTATE: { type: GraphQLString },
        ENROLLMENTSOURCE: { type: GraphQLString },
        ADDRESSES: {
            type: new GraphQLList(Address),
            resolve(obj, args, ast){
                return Resolver(args.id).Address();
        }}
    })
});

这是我的地址对象，由第二个 REST 调用解析：

let Address = new GraphQLObjectType({
    name: 'ADDRESS',
    fields: () => ({
        ACTIVE: { type: GraphQLString },
        ADDRESS1: { type: GraphQLString },
        ADDRESS2: { type: GraphQLString },
        ADDRESS3: { type: GraphQLString },
        CAMPAIGN: { type: GraphQLString },
        CITY: { type: GraphQLString },
        STATE: { type: GraphQLString },
        STATUS: { type: GraphQLString },
        TIMECREATED: { type: GraphQLString },
        TYPE: { type: GraphQLString },
        ZIP: { type: GraphQLString },
    })

});

export default Address;

这些是我的解析器

var Resolver = (id) => {

    var options = {
        hostname: "myhostname",
        port: 4000
    };


    var GetPromise = (options, id, path) => {
        return new Promise((resolve, reject) => {
            http.get(options, (response) => {
                var completeResponse = '';
                response.on('data', (chunk) => {
                    completeResponse += chunk;
                });
                response.on('end', () => {
                    parser.parseString(completeResponse, (err, result) => {
                        let pathElements = path.split('.');                       
                        resolve(result[pathElements[0]][pathElements[1]]);
                    });
                });
            }).on('error', (e) => { });
        });
    };

    let Student= () => {
        options.path = '/Student/' + id;
        return GetPromise(options, id, 'GetStudentResult.StudentINFO');
    }

    let Address= () => {
         options.path = '/Address/' + id + '/All';
        return GetPromise(options, id, 'getAddressResult.ADDRESS');
    };


    return {
        Student,
        Address
    };
}

export default Resolver;

Answer 1

ADDRESSES: {
    type: new GraphQLList(Address),
    resolve(obj, args, ast){
        return Resolver(args.id).Address();
    }
}

传递给 ADDRESSES 的参数是在查询时传递给 ADDRESSES 字段的参数。在 resolve 方法中，obj 应该是学生对象，如果你有一个 id 属性，你需要做的就是：return Resolver(obj.id).Address();。

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