如何在GraphQL中进行简单的联接? [英] How to do a simple join in GraphQL?

问题描述

我是GraphQL的新手，正在尝试做一个简单的联接查询.我的示例表如下所示:

I am very new in GraphQL and trying to do a simple join query. My sample tables look like below:

{
  phones: [
    {
      id: 1,
      brand: 'b1',
      model: 'Galaxy S9 Plus',
      price: 1000,
    },
    {
      id: 2,
      brand: 'b2',
      model: 'OnePlus 6',
      price: 900,
    },
  ],
  brands: [
    {
      id: 'b1',
      name: 'Samsung'
    },
    {
      id: 'b2',
      name: 'OnePlus'
    }
  ]
}

我想查询一个返回 phone 对象，其中带有其品牌名称而不是品牌代码.

I would like to have a query to return a phone object with its brand name in it instead of the brand code.

例如如果使用 id = 2 查询电话，它将返回:

E.g. If queried for the phone with id = 2, it should return:

{id: 2, brand: 'OnePlus', model: 'OnePlus 6', price: 900}

推荐答案

GraphQL作为前端的查询语言，在传统的SQL意义上不支持联接".

GraphQL as a query language on the front-end does not support 'joins' in the classic SQL sense.

相反，它允许您选择要为组件获取的特定模型中的哪些字段.

Rather, it allows you to pick and choose which fields in a particular model you want to fetch for your component.

要查询数据集中的所有电话，您的查询应如下所示:

To query all phones in your dataset, your query would look like this:

query myComponentQuery {
  phone {
    id
    brand
    model
    price
  }
}

您的前端正在查询的GraphQL服务器将具有单独的字段解析器-告诉GraphQL在哪里获取ID，品牌，型号等.

The GraphQL server that your front-end is querying would then have individual field resolvers - telling GraphQL where to fetch id, brand, model etc.

服务器端解析器如下所示:

The server-side resolver would look something like this:

Phone: {
  id(root, args, context) {
    pg.query('Select * from Phones where name = ?', ['blah']).then(d => {/*doStuff*/})
    //OR
    fetch(context.upstream_url + '/thing/' + args.id).then(d => {/*doStuff*/})

    return {/*the result of either of those calls here*/}
  },
  price(root, args, context) {
    return 9001
  },
},

这篇关于如何在GraphQL中进行简单的联接?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！