Relay / GraphQL 'resolve'的工作原理

How does Relay / GraphQL 'resolve' works?

本文关键字：工作 resolve Relay GraphQL      更新时间：2023-09-26

我正在尝试Relay和GraphQL。当我在做模式时，我这样做:

let articleQLO = new GraphQLObjectType({
  name: 'Article',
  description: 'An article',
  fields: () => ({
    _id: globalIdField('Article'),
    title: {
      type: GraphQLString,
      description: 'The title of the article',
      resolve: (article) => article.getTitle(),
    },
    author: {
      type: userConnection,
      description: 'The author of the article',
      resolve: (article) => article.getAuthor(),
    },
  }),
  interfaces: [nodeInterface],
})

所以，当我要求一篇这样的文章时:

{
  article(id: 1) {
    id,
    title,
    author
  }
}

它会做3查询数据库?我的意思是，每个字段都有一个解析方法(getTitle, getAuthor等)，它向数据库发出请求。我做错了吗?

这是getAuthor的一个例子(我用猫鼬):

articleSchema.methods.getAuthor = function(id){
  let article = this.model('Article').findOne({_id: id})
  return article.author
}

如果resolve方法被传递给article，你不能直接访问属性吗?

let articleQLO = new GraphQLObjectType({
  name: 'Article',
  description: 'An article',
  fields: () => ({
    _id: globalIdField('Article'),
    title: {
      type: GraphQLString,
      description: 'The title of the article',
      resolve: (article) => article.title,
    },
    author: {
      type: userConnection,
      description: 'The author of the article',
      resolve: (article) => article.author,
    },
  }),
  interfaces: [nodeInterface],
})

由于Mongoose中的Schema.methods在模型上定义了方法，因此它不需要文章的ID(因为您在文章实例上调用它)。所以，如果你想保留这个方法，你只需要:

articleSchema.methods.getAuthor = function() {
  return article.author;
}

如果它是你需要查找的东西，例如在另一个集合中，然后你需要做一个单独的查询(假设你不使用refs):

articleSchema.methods.getAuthor = function(callback) {
  return this.model('Author').find({ _id: this.author_id }, cb);
}
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