Reputation: 759
I have just learnt GraphQL and I want to find the user id=2 OR user id=3 now how will I make the GraphQL query,I am getting the whole set using the bellow query
{
users() {
id
username
posts {
title
tags {
name
}
}
}
}
2nd issue --
{
people(id:[1,2,3]) {
id
username
posts(id:2) {
title
tags {
name
}
}
}
}
if I add the arg on Posts filed then i got an error msg "Unknown argument id on field posts of type user"
This is my Schema js file
var graphql = require('graphql');
var Db = require('./db');
var users = new graphql.GraphQLObjectType({
name : 'user',
description : 'this is user info',
fields : function(){
return {
id :{
type : graphql.GraphQLInt,
resolve(user){
return user.id;
}
},
username :{
type : graphql.GraphQLString,
resolve(user){
return user.username;
}
},
posts:{
id:{
type : graphql.GraphQLString,
resolve(post){
return post.id;
}
},
type: new graphql.GraphQLList(posts),
resolve(user){
return user.getPosts();
}
}
}
}
});
var posts = new graphql.GraphQLObjectType({
name : 'Posts',
description : 'this is post info',
fields : function(){
return {
id :{
type : graphql.GraphQLInt,
resolve(post){
return post.id;
}
},
title :{
type : graphql.GraphQLString,
resolve(post){
return post.title;
}
},
content:{
type : graphql.GraphQLString,
resolve(post){
return post.content;
}
},
person :{
type: users,
resolve(post){
return post.getUser();
}
},
tags :{
type: new graphql.GraphQLList(tags),
resolve(post){
return post.getTags();
}
}
}
}
});
var tags = new graphql.GraphQLObjectType({
name : 'Tags',
description : 'this is Tags info',
fields : function(){
return {
id :{
type : graphql.GraphQLInt,
resolve(tag){
return tag.id;
}
},
name:{
type : graphql.GraphQLString,
resolve(tag){
return tag.name;
}
},
posts :{
type: new graphql.GraphQLList(posts),
resolve(tag){
return tag.getPosts();
}
}
}
}
});
var query = new graphql.GraphQLObjectType({
name : 'query',
description : 'Root query',
fields : function(){
return {
people :{
type : new graphql.GraphQLList(users),
args :{
id:{type: new graphql.GraphQLList(graphql.GraphQLInt)},
username:{
type: graphql.GraphQLString
}
},
resolve(root,args){
return Db.models.user.findAll({where:args});
}
},
posts:{
type : new graphql.GraphQLList(posts),
args :{
id:{
type: graphql.GraphQLInt
},
title:{
type: graphql.GraphQLString
},
},
resolve(root,args){
return Db.models.post.findAll({where:args});
}
},
tags :{
type : new graphql.GraphQLList(tags),
args :{
id:{
type: graphql.GraphQLInt
},
name:{
type: graphql.GraphQLString
},
},
resolve(root,args){
return Db.models.tag.findAll({where:args});
}
}
}
}
});
var Mutation = new graphql.GraphQLObjectType({
name : "mutation",
description : 'function for mutaion',
fields : function(){
return {
addPerson : {
type : users,
args :{
username : {
type : new graphql.GraphQLNonNull(graphql.GraphQLString)
},
email :{
type : new graphql.GraphQLNonNull(graphql.GraphQLString)
}
},
resolve(_, args){
return Db.models.user.create({
username : args.username,
email : args.email
});
}
}
}
}
})
var Schama = new graphql.GraphQLSchema({
query : query,
mutation : Mutation
})
module.exports = Schama;
Upvotes: 0
Views: 742
Reputation: 2245
In order to fetch multiple data from your schema using an array of id
s you should define the args given to the users
in your schema as follows:
fields: () => ({
users: {
type: new GraphQLList(USER_GRAPHQL_OBJECT_TYPE),
args: {
id: {type: new GraphQLList(GraphQLInt)}
},
resolve: (root, args) => {
// fetch users
}
}
})
notice the new GraphQLList
wrapping the GraphQLInt
type of the id.
Then, when querying your schema you can :
{
users(id: [2, 3]) {
id
username
posts {
title
tags {
name
}
}
}
}
Please let me know if it was helpful :)
Upvotes: 1