GraphQL：如何将参数传递给子对象

问题描述：

我正在使用GraphQL来查询将由大约15个不同的REST调用组成的对象。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。我尝试向地址对象添加参数，但是我得到一个错误，指示参数不从Student对象传递。所以我的问题是：我如何将客户端查询中的参数传递给GraphQL服务器中的子对象？GraphQL：如何将参数传递给子对象

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;

这是我的主要学生对象，我链接其他对象。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(); 
     }} 
    }) 
});

这里是由第二REST调用解决了我的地址对象：

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;

这些都是我的解析器

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;

FallFast：你找到任何解决方案 – Brune

你找到一个好的解决方案吗？ – James111

答

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

传递至ADDRESSES

ARG游戏参数传递给ADDRESSES字段在查询时间。在解决方法中，obj应该是学生对象，如果您有id属性，则只需要执行以下操作：return Resolver(obj.id).Address();。

GraphQL：如何将参数传递给子对象

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