如何从与主表相同级别的续集中的连接表返回值

Question

我正在尝试找到一种方法，如何将所有连接表置于与我的主表相同的级别...到目前为止，它只会导致我的最终对象中的嵌套值..

这是我所拥有的

Orders.findAll({
   include: [
      {model: Products, attributes: ['product_name']}
   ],
   attributes: ['id_order', 'dtime_order', 'amount']
})

我得到的是：

[
 {
   id_order: 1, 
   dtime_order: '2021-05-24T22:00:00.000Z',
   amount: 20,
   products: {
      product_name: 'Picture'
   }
 }
]

但我想要得到的是：

[
 {
   id_order: 1, 
   dtime_order: '2021-05-24T22:00:00.000Z',
   amount: 20,
   product_name: 'Picture'
 }
]

我尝试过这个How to return result from include model in same level of main model in Sequelize? 但不幸的是当我这样做时：

Orders.findAll({
   include: [
      {model: Products, attributes: []}
   ],
   attributes: ['id_order', 'dtime_order', 'amount', ['products.product_name', 'product_name']]
})

对我来说不起作用

column "products.product_name" does not exist

在将对象发送回响应之前可能有一种奇怪的方法来修改对象..但我宁愿在 Sequelize 中进行..

任何想法都非常受欢迎...谢谢你们！

编辑：添加生成的 SQL

Executing (default): SELECT "orders"."id_order", "orders"."dtime_order", "orders"."amount", "products.product_name", FROM "orders" AS "orders" LEFT OUTER JOIN "products" AS "products" ON "orders"."id_order" = "products"."ir_order";
error:   Get dashboard data error: column "products.product_name" does not exist

解决方案：

我必须在我的协会中使用别名

Orders.hasOne(Products, {as: 'products', ....})

然后在我的包含和引用中使用完全相同的别名

include: [{model: Products, attributes: [], as: 'products'}]

和

attributes: [ ... , [Sequelize.col('products.product_name', 'product_name')]

没有像raw: true魅力一样的作品:) 谢谢@Emma !!!

Answer 1

请使用Sequelize.col来包装嵌套列，以便 Sequelize 可以正确地为该列添加别名。

Orders.findAll({
  include: [
    {model: Products, attributes: []}
  ],
  attributes: ['id_order', 'dtime_order', 'amount', [Sequelize.col('products.product_name'), 'product_name']],
  raw: true
})