Ecmascript 6 GraphQL中的自定义类型
我正在使用。我想在其中实现一对多关联。我有两种类型用户和办公室。一个用户有多个办公室 用户类型:Ecmascript 6 GraphQL中的自定义类型,ecmascript-6,graphql,graphql-js,Ecmascript 6,Graphql,Graphql Js,我正在使用。我想在其中实现一对多关联。我有两种类型用户和办公室。一个用户有多个办公室 用户类型: var graphql = require('graphql'); const userType = new graphql.GraphQLObjectType({ name: 'user', fields :()=>{ var officeType=require('./officeSchema'); return { _id: { type: graphql.GraphQLID
var graphql = require('graphql');
const userType = new graphql.GraphQLObjectType({
name: 'user',
fields :()=>{
var officeType=require('./officeSchema');
return {
_id: {
type: graphql.GraphQLID
},
name: {
type: graphql.GraphQLString
},
age: {
type: graphql.GraphQLString
},
office:{
type:officeType
}
};
}
});
module.exports=userType;
const officeInputType = new graphql.GraphQLInputObjectType({
name: 'officeinput',
fields: () => {
return {
room: {
type: graphql.GraphQLString
},
location: {
type: graphql.GraphQLString
}
}
}
});
const userInputType = new graphql.GraphQLInputObjectType({
name: 'userinput',
fields: () => {
return {
name: {
type: graphql.GraphQLString
},
age: {
type: graphql.GraphQLString
}
}
}
});
办公模式:
const officeType = new graphql.GraphQLObjectType({
name: 'office',
fields:()=> {
var userType = require('./userSchema');
return {
_id: {
type: graphql.GraphQLID
},
room: {
type: graphql.GraphQLString
},
location: {
type: graphql.GraphQLString
},
users: {
type: new graphql.GraphQLList(userType),
resolve: (obj,{_id}) => {
fetch('http://0.0.0.0:8082/office/user/'+obj._id, {
method: "GET",
headers: {
'Content-Type': 'application/json'
}
})
.then(function(res) {return res});
}
}
};
}
});
现在变异代码如下:
const Adduser = {
type: userType,
args: {
name: {
type: graphql.GraphQLString
},
age: {
type: graphql.GraphQLString
}
},
resolve: (obj, {
input
}) => {
}
};
const Addoffice = {
type: OfficeType,
args: {
room: {
type: graphql.GraphQLString
},
location: {
type: graphql.GraphQLString
},
users: {
type: new graphql.GraphQLList(userInputType)
}
},
resolve: (obj, {
input
}) => {
}
};
const Rootmutation = new graphql.GraphQLObjectType({
name: 'Rootmutation',
fields: {
Adduser: Adduser,
Addoffice: Addoffice
}
});
此代码正在抛出错误,如下所示
Rootmutation.Addoffice(用户:)参数类型必须是输入类型,但得到的却是:[user]。
我想在数据库中添加实际字段以及相关表的字段,但无法解决问题
更新:
var graphql = require('graphql');
const userType = new graphql.GraphQLObjectType({
name: 'user',
fields :()=>{
var officeType=require('./officeSchema');
return {
_id: {
type: graphql.GraphQLID
},
name: {
type: graphql.GraphQLString
},
age: {
type: graphql.GraphQLString
},
office:{
type:officeType
}
};
}
});
module.exports=userType;
const officeInputType = new graphql.GraphQLInputObjectType({
name: 'officeinput',
fields: () => {
return {
room: {
type: graphql.GraphQLString
},
location: {
type: graphql.GraphQLString
}
}
}
});
const userInputType = new graphql.GraphQLInputObjectType({
name: 'userinput',
fields: () => {
return {
name: {
type: graphql.GraphQLString
},
age: {
type: graphql.GraphQLString
}
}
}
});
1-添加的GraphQlinputObject类型:
var graphql = require('graphql');
const userType = new graphql.GraphQLObjectType({
name: 'user',
fields :()=>{
var officeType=require('./officeSchema');
return {
_id: {
type: graphql.GraphQLID
},
name: {
type: graphql.GraphQLString
},
age: {
type: graphql.GraphQLString
},
office:{
type:officeType
}
};
}
});
module.exports=userType;
const officeInputType = new graphql.GraphQLInputObjectType({
name: 'officeinput',
fields: () => {
return {
room: {
type: graphql.GraphQLString
},
location: {
type: graphql.GraphQLString
}
}
}
});
const userInputType = new graphql.GraphQLInputObjectType({
name: 'userinput',
fields: () => {
return {
name: {
type: graphql.GraphQLString
},
age: {
type: graphql.GraphQLString
}
}
}
});
2-在AddOffice中添加了userinputtype而不是usertype。
现在错误是
Rootmutation.Addoffice(user:) argument type must be Input Type but got: userinput.
问题是您提供了
userType
作为Addoffice
变量类型之一userType
不能是参数类型。相反,您必须使用输入类型
有两种对象类型:输出类型和输入类型。您的userType
和officeType
是输出类型。您需要使用GraphQLInputObjectType
[]创建输入类型。它可能会有非常相似的领域。您可以将其用作参数字段的类型
const userInputType = new graphql.GraphQLInputObjectType({
name: 'UserInput',
fields () => {
return {
_id: {
type: graphql.GraphQLID
},
// ...
};
}
});
根据你的建议更改了代码。请看看我现在做错了什么?