Graphql字段上的未知参数

Question

我是GraphQL的新手，当我尝试发送args on（posts）子节点时，如bellow查询我收到错误消息＆＃34;未知参数id在类型user＆＃34;的字段帖子上。我想带一些特定的帖子不是全部。

{ people(id:[1,2]) {
            id
            username
            posts(id:2) {
              title
              tags {
                name
              }
            }
          }
        }

这是我的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 Schama = new graphql.GraphQLSchema({
  query : query,
  mutation : Mutation
})

module.exports = Schama;

Answer 1

看起来你缺少用户的args，因此，它应该是这样的：

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:{
        args: {
        id:{
            type : graphql.GraphQLInt,
        },
        type: new  graphql.GraphQLList(posts),
        resolve(user, args){
          // Code here to use args.id
          return user.getPosts();
        }
      }


    }
  }
});

Answer 2

如果参数在字段中不存在，则会出现此错误。

例如，在这种情况下，列表中不存在invalidArg。唯一可能的参数是lastB，之后，第一个，之前，orderBy和ownedByViewer。如果您尝试传递其他任何内容，都会收到错误消息。

Answer 3

尝试用 posts（单数）替换 post（复数）。

post(id:2) {
          title
          tags {
            name
          }
        }