[英]GraphQL, Mysql equivalent OR operation
我剛剛學習了GraphQL,現在我想找到用戶ID = 2 或用戶ID = 3,我將如何進行GraphQL查詢,使用波紋管查詢獲取整個集合
{
users() {
id
username
posts {
title
tags {
name
}
}
}
}
第二期-
{
people(id:[1,2,3]) {
id
username
posts(id:2) {
title
tags {
name
}
}
}
}
如果我在提交的帖子上添加arg,那么我會收到一條錯誤消息:“類型為user的字段帖子上的未知參數ID”
這是我的Schema js文件
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;
為了使用id
數組從架構中獲取多個數據,您應該定義在架構中為users
提供的arg,如下所示:
fields: () => ({
users: {
type: new GraphQLList(USER_GRAPHQL_OBJECT_TYPE),
args: {
id: {type: new GraphQLList(GraphQLInt)}
},
resolve: (root, args) => {
// fetch users
}
}
})
請注意, new GraphQLList
包裝了id的GraphQLInt
類型。
然后,在查詢架構時,您可以:
{
users(id: [2, 3]) {
id
username
posts {
title
tags {
name
}
}
}
}
請讓我知道是否有幫助:)
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