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javascript - Relay/GraphQL 'resolve' 是如何工作的?

转载 作者:数据小太阳 更新时间:2023-10-29 04:45:53 28 4
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我正在尝试 RelayGraphQL .当我在做模式时,我是这样做的:

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,当我要求这样一篇文章时:

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

它会对数据库进行 3 次查询吗?我的意思是,每个字段都有一个向数据库发出请求的解析方法(getTitlegetAuthor 等)。我做错了吗?

这是一个 getAuthor 的例子(我用的是 mongoose):

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

最佳答案

如果 resolve 方法被传递给 article,您不能只访问该属性吗?

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,你只需要做:

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

如果这是您需要查找的内容,例如在另一个集合中,然后你需要做一个单独的查询(假设你没有使用 refs):

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

关于javascript - Relay/GraphQL 'resolve' 是如何工作的?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32226857/

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