如何使用GraphQL和node实现接口
How to implement interface using GraphQL and node
我想在另一种对象类型中实现一种对象类型的字段
这是我的架构文件。
const Films = new GraphQLInterfaceType({
name: 'films',
fields: () => ({
id:{
type: GraphQLID
},
name: {
type: GraphQLString,
},
})
})
const MovieStream = new GraphQLObjectType({
name: 'MovieStream',
interfaces: () => [Films],
fields: () => ({
id: {
type: GraphQLID,
},
movie_id: {
type: GraphQLString,
},
})
})
这里我尝试使用界面。但它显示错误:
{
"errors": [
{
"message": "Query root type must be Object type, it cannot be { __validationErrors: undefined, __allowedLegacyNames: [], _queryType: undefined, _mutationType: undefined, _subscriptionType: undefined, _directives: [@include, @skip, @deprecated], astNode: undefined, extensionASTNodes: undefined, _typeMap: { __Schema: __Schema, __Type: __Type, __TypeKind: __TypeKind, String: String, Boolean: Boolean, __Field: __Field, __InputValue: __InputValue, __EnumValue: __EnumValue, __Directive: __Directive, __DirectiveLocation: __DirectiveLocation, films: films, ID: ID, Date: Date, JSON: JSON, MovieStream: MovieStream }, _possibleTypeMap: {}, _implementations: { films: [] } }."
},
{
"message": "Expected GraphQL named type but got: { __validationErrors: undefined, __allowedLegacyNames: [], _queryType: undefined, _mutationType: undefined, _subscriptionType: undefined, _directives: [@include, @skip, @deprecated], astNode: undefined, extensionASTNodes: undefined, _typeMap: { __Schema: __Schema, __Type: __Type, __TypeKind: __TypeKind, String: String, Boolean: Boolean, __Field: __Field, __InputValue: __InputValue, __EnumValue: __EnumValue, __Directive: __Directive, __DirectiveLocation: __DirectiveLocation, films: films, ID: ID, Date: Date, JSON: JSON, MovieStream: MovieStream }, _possibleTypeMap: {}, _implementations: { films: [] } }."
}
]
}
这是查询类型:
const QueryRoot = new GraphQLObjectType({
name: 'Query',
fields: () => ({
getContentList:{
type: new GraphQLList(contentCategory),
args: {
id: {
type: GraphQLInt
},
permalink: {
type: GraphQLString
},
language: {
type: GraphQLString
},
content_types_id: {
type: GraphQLString
},
oauth_token:{
type: GraphQLString
}
},
resolve: (parent, args, context, resolveInfo) => {
var category_flag = 0;
var menuItemInfo = '';
user_id = args.user_id ? args.user_id : 0;
// console.log("context"+context['oauth_token']);
return AuthDb.models.oauth_registration.findAll({attributes: ['oauth_token', 'studio_id'],where:{
// oauth_token:context['oauth_token'],
$or: [
{
oauth_token:
{
$eq: context['oauth_token']
}
},
{
oauth_token:
{
$eq: args.oauth_token
}
},
]
},limit:1}).then(oauth_registration => {
var oauthRegistration = oauth_registration[0]
// for(var i = 0;i<=oauth_registration.ength;i++){
if(oauth_registration && oauthRegistration && oauthRegistration.oauth_token == context['oauth_token'] || oauthRegistration.oauth_token == args.oauth_token){
studio_id = oauthRegistration.studio_id;
return joinMonster.default(resolveInfo,{}, sql => {
return contentCategoryDb.query(sql).then(function(result) {
return result[0];
});
} ,{dialect: 'mysql'});
}else{
throw new Error('Invalid OAuth Token');
}
})
},
where: (filmTable, args, context) => {
return getLanguage_id(args.language).then(language_id=>{
return ` ${filmTable}.permalink = "${args.permalink}" and ${filmTable}.studio_id = "${studio_id}" and (${filmTable}.language_id = "${language_id}" OR ${filmTable}.parent_id = 0 AND ${filmTable}.id NOT IN (SELECT ${filmTable}.parent_id FROM content_category WHERE ${filmTable}.permalink = "${args.permalink}" and ${filmTable}.language_id = "${language_id}" and ${filmTable}.studio_id = "${studio_id}"))`
})
},
}
})
})
module.exports = new GraphQLSchema({
query: QueryRoot
})
请帮帮我。我是不是在界面的使用上做错了什么?
我通过这个找到了答案post
Is it possible to fetch data from multiple tables using GraphQLList
任何人请告诉我在我的代码中使用接口的确切方法。
尽管您打印的错误与 interfaces
实现没有真正相关,但为了使用接口,您必须实现 methods/types 接口引用。因此,在您的情况下,您的对象 MovieStream
缺少您在对象 Films
中引用的类型 name
。
您的代码应该类似于:
const Films = new GraphQLInterfaceType({
name: 'films',
fields: () => ({
id:{
type: GraphQLID
},
name: {
type: GraphQLString,
},
})
})
const MovieStream = new GraphQLObjectType({
name: 'MovieStream',
interfaces: () => [Films],
fields: () => ({
id: {
type: GraphQLID,
},
name: {
type: GraphQLString // You're missing this!
},
movie_id: {
type: GraphQLString,
},
})
})
现在回到你打印的错误"message": "Query root type must be Object type, it cannot be...
这好像和你的QueryRoot
对象有关,好像GraphQLSchema
没有识别根对象。如果修复界面后此问题仍然存在,请查看此答案
我想在另一种对象类型中实现一种对象类型的字段
这是我的架构文件。
const Films = new GraphQLInterfaceType({
name: 'films',
fields: () => ({
id:{
type: GraphQLID
},
name: {
type: GraphQLString,
},
})
})
const MovieStream = new GraphQLObjectType({
name: 'MovieStream',
interfaces: () => [Films],
fields: () => ({
id: {
type: GraphQLID,
},
movie_id: {
type: GraphQLString,
},
})
})
这里我尝试使用界面。但它显示错误:
{
"errors": [
{
"message": "Query root type must be Object type, it cannot be { __validationErrors: undefined, __allowedLegacyNames: [], _queryType: undefined, _mutationType: undefined, _subscriptionType: undefined, _directives: [@include, @skip, @deprecated], astNode: undefined, extensionASTNodes: undefined, _typeMap: { __Schema: __Schema, __Type: __Type, __TypeKind: __TypeKind, String: String, Boolean: Boolean, __Field: __Field, __InputValue: __InputValue, __EnumValue: __EnumValue, __Directive: __Directive, __DirectiveLocation: __DirectiveLocation, films: films, ID: ID, Date: Date, JSON: JSON, MovieStream: MovieStream }, _possibleTypeMap: {}, _implementations: { films: [] } }."
},
{
"message": "Expected GraphQL named type but got: { __validationErrors: undefined, __allowedLegacyNames: [], _queryType: undefined, _mutationType: undefined, _subscriptionType: undefined, _directives: [@include, @skip, @deprecated], astNode: undefined, extensionASTNodes: undefined, _typeMap: { __Schema: __Schema, __Type: __Type, __TypeKind: __TypeKind, String: String, Boolean: Boolean, __Field: __Field, __InputValue: __InputValue, __EnumValue: __EnumValue, __Directive: __Directive, __DirectiveLocation: __DirectiveLocation, films: films, ID: ID, Date: Date, JSON: JSON, MovieStream: MovieStream }, _possibleTypeMap: {}, _implementations: { films: [] } }."
}
]
}
这是查询类型:
const QueryRoot = new GraphQLObjectType({
name: 'Query',
fields: () => ({
getContentList:{
type: new GraphQLList(contentCategory),
args: {
id: {
type: GraphQLInt
},
permalink: {
type: GraphQLString
},
language: {
type: GraphQLString
},
content_types_id: {
type: GraphQLString
},
oauth_token:{
type: GraphQLString
}
},
resolve: (parent, args, context, resolveInfo) => {
var category_flag = 0;
var menuItemInfo = '';
user_id = args.user_id ? args.user_id : 0;
// console.log("context"+context['oauth_token']);
return AuthDb.models.oauth_registration.findAll({attributes: ['oauth_token', 'studio_id'],where:{
// oauth_token:context['oauth_token'],
$or: [
{
oauth_token:
{
$eq: context['oauth_token']
}
},
{
oauth_token:
{
$eq: args.oauth_token
}
},
]
},limit:1}).then(oauth_registration => {
var oauthRegistration = oauth_registration[0]
// for(var i = 0;i<=oauth_registration.ength;i++){
if(oauth_registration && oauthRegistration && oauthRegistration.oauth_token == context['oauth_token'] || oauthRegistration.oauth_token == args.oauth_token){
studio_id = oauthRegistration.studio_id;
return joinMonster.default(resolveInfo,{}, sql => {
return contentCategoryDb.query(sql).then(function(result) {
return result[0];
});
} ,{dialect: 'mysql'});
}else{
throw new Error('Invalid OAuth Token');
}
})
},
where: (filmTable, args, context) => {
return getLanguage_id(args.language).then(language_id=>{
return ` ${filmTable}.permalink = "${args.permalink}" and ${filmTable}.studio_id = "${studio_id}" and (${filmTable}.language_id = "${language_id}" OR ${filmTable}.parent_id = 0 AND ${filmTable}.id NOT IN (SELECT ${filmTable}.parent_id FROM content_category WHERE ${filmTable}.permalink = "${args.permalink}" and ${filmTable}.language_id = "${language_id}" and ${filmTable}.studio_id = "${studio_id}"))`
})
},
}
})
})
module.exports = new GraphQLSchema({
query: QueryRoot
})
请帮帮我。我是不是在界面的使用上做错了什么?
我通过这个找到了答案post Is it possible to fetch data from multiple tables using GraphQLList
任何人请告诉我在我的代码中使用接口的确切方法。
尽管您打印的错误与 interfaces
实现没有真正相关,但为了使用接口,您必须实现 methods/types 接口引用。因此,在您的情况下,您的对象 MovieStream
缺少您在对象 Films
中引用的类型 name
。
您的代码应该类似于:
const Films = new GraphQLInterfaceType({
name: 'films',
fields: () => ({
id:{
type: GraphQLID
},
name: {
type: GraphQLString,
},
})
})
const MovieStream = new GraphQLObjectType({
name: 'MovieStream',
interfaces: () => [Films],
fields: () => ({
id: {
type: GraphQLID,
},
name: {
type: GraphQLString // You're missing this!
},
movie_id: {
type: GraphQLString,
},
})
})
现在回到你打印的错误"message": "Query root type must be Object type, it cannot be...
这好像和你的QueryRoot
对象有关,好像GraphQLSchema
没有识别根对象。如果修复界面后此问题仍然存在,请查看此答案