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javascript - GraphQL、Mysql 等价 OR 运算

转载 作者:行者123 更新时间:2023-12-03 06:02:42 29 4
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我刚刚学习了 GraphQL,我想找到用户 id=2 OR 用户 id=3 现在我将如何进行 GraphQL 查询,我正在使用以下查询获取整个集合

 {
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参数,则会收到错误消息“用户类型的字段帖子上的参数 id 未知”

这是我的架构 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;

最佳答案

为了使用 id 数组从架构中获取多个数据,您应该定义为架构中的 user 提供的参数,如下所示:

fields: () => ({
users: {
type: new GraphQLList(USER_GRAPHQL_OBJECT_TYPE),
args: {
id: {type: new GraphQLList(GraphQLInt)}
},
resolve: (root, args) => {
// fetch users
}
}
})

注意new GraphQLList包装了id的GraphQLInt类型。

然后,在查询您的架构时,您可以:

{
users(id: [2, 3]) {
id
username
posts {
title
tags {
name
}
}
}
}

请告诉我是否有帮助:)

关于javascript - GraphQL、Mysql 等价 OR 运算,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39690413/

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