错误:Note.user字段类型必须是输出类型,但得到:[对象对象]
问题描述:
我对graphql非常新,我正在玩一个简单的案例,不能使它的工作,案件是一对多,用户有很多音符,一个音符属于用户。 我的问题是与“NoteType”我不能返回“用户类型”,我可以返回创建的注意,但不是用户类型错误:Note.user字段类型必须是输出类型,但得到:[对象对象]
代码的用户的用户名(重要删除某些字段)
NoteType
当我取消去用户我有错误错误:Note.user字段类型必须是输出类型,但得到:[对象OBJE克拉。
import {
GraphQLObjectType,
GraphQLInt,
GraphQLString,
GraphQLBoolean,
} from 'graphql';
import UserType from '../User/UserType';
import User from '../User/User';
const NoteType = new GraphQLObjectType({
name: 'Note',
description: 'This represents a Note',
fields:() => ({
id: {
type: GraphQLInt,
resolve: (note) => note.id,
},
userId: {
type: GraphQLInt,
resolve: (note) => note.userId,
},
title: {
type: GraphQLString,
resolve: (note) => note.note,
},
username: {
type: GraphQLString,
resolve: (note) => (
User
.findOne({
where: {
id: note.userId,
},
}).then(user => user.username)
),
},
/* user: { ***PROBLEM HERE!!***
type: UserType,
resolve: (note) => (
User
.findOne({
where: {
id: note.userId,
},
}).then(user => user)
),
},*/
}),
});
module.exports = NoteType;
的UserType
import {
GraphQLObjectType,
GraphQLInt,
GraphQLString,
GraphQLList,
} from 'graphql';
import NoteType from '../Note/NoteType';
import Note from '../Note/Note';
import LocationType from '../Location/LocationType';
import Location from '../Location/Location';
const UserType = new GraphQLObjectType({
name: 'User',
description: 'This represents a User',
fields:() => ({
id: {
type: GraphQLInt,
resolve: (user) => user.id,
},
username: {
type: GraphQLString,
resolve: (user) => user.username,
},
email: {
type: GraphQLString,
resolve: (user) => user.email,
},
notes: {
type: new GraphQLList(NoteType),
resolve: (user) => (
Note
.findAll({
where: {
userId: user.id,
},
})
),
},
}),
});
module.exports = UserType;
NoteQuery
import {
GraphQLInt,
GraphQLString,
GraphQLList,
} from 'graphql';
import NoteType from '../../models/Note/NoteType';
import Note from '../../models/Note/Note';
import UserType from '../../models/User/UserType';
import User from '../../models/User/User';
const noteQuery = {
type: new GraphQLList(NoteType),
args: {
id: {
name: 'id',
type: GraphQLInt,
},
userId: {
name: 'userId',
type: GraphQLInt,
},
user: {
name: 'user',
type: GraphQLString,
},
note: {
name: 'note',
type: GraphQLString,
},
createdAt: {
name: 'createdAt',
type: GraphQLString,
},
updatedAt: {
name: 'updatedAt',
type: GraphQLString,
},
},
resolve: (user, args) => Note.findAll({ where: args }),
};
module.exports = noteQuery;
UserQuery
import {
GraphQLInt,
GraphQLString,
GraphQLList,
} from 'graphql';
import UserType from '../../models/User/UserType';
import User from '../../models/User/User';
const userQuery = {
users: {
type: new GraphQLList(UserType),
args: {
id: {
name: 'id',
type: GraphQLInt,
},
username: {
name: 'username',
type: GraphQLString,
},
email: {
name: 'email',
type: GraphQLString,
},
createdAt: {
name: 'createdAt',
type: GraphQLString,
},
updatedAt: {
name: 'updatedAt',
type: GraphQLString,
},
},
resolve: (user, args) => User.findAll({ where: args }),
},
user: {
type: UserType,
args: {
id: {
name: 'id',
type: GraphQLInt,
},
username: {
name: 'username',
type: GraphQLString,
},
email: {
name: 'email',
type: GraphQLString,
},
createdAt: {
name: 'createdAt',
type: GraphQLString,
},
updatedAt: {
name: 'updatedAt',
type: GraphQLString,
},
},
resolve: (user, args) => User.findOne({ where: args }),
},
};
module.exports = userQuery;
任何帮助或提示,提前致谢!
答
看起来像循环依赖性问题。您将用户模块导入Note模块,同时将Note模块导入您的用户模块。 user
字段的解析器正在查找GraphQLObjectType的实例,但它只接收导出类型的未完成副本。
您可以查看this question以获得更全面的解释和一些解决方法。
但是,我建议只需使用buildSchema并使用解析器通过根对象传递字符串文本来声明模式。或者使用graphql-tool
的makeExecutableSchema,这更容易。无论哪种方式,您都可以节省处理循环依赖的头痛问题,并使您的模式更具可读性。
谢谢,我会调查,谢谢你的提示! – Julieta