javascript - 如何解决嵌套的promise？

Question

我正在尝试解决分散在数组内对象中的多个 promise 。这是我的代码的简单版本:

const getOrderSum = order_id => {
    return new Promise(resolve => {
        setTimeout(() => resolve(34), 1000)
    })
}

const customer_array = [
    {
        customer_id: '19847743234730384',
        name: 'Customer 1',
        orders: [
            {
                order_id: '98749873244324',
                price_per: 12
            },
            {
                order_id: '9874987323545',
                price_per: 16
            }
        ]
    },
    {
        customer_id: '123454351234123',
        name: 'Customer 2',
        orders: [
            {
                order_id: '918741433423',
                price_per: 6
            }
        ]
    }
]

const result_array = customer_array.map(customer => {
    const promises = customer.orders.map(async order => {
        order.order_total = await getOrderSum();
        return order;
    });

    customer.orders = promises;
    return customer;
})

console.log(result_array);

在 map 循环内，我想调用一个异步函数，该函数为每个订单返回一个 order_total 。代码运行后 result_array 看起来像这样:

[
  {
    customer_id: '19847743234730384',
    name: 'Customer 1',
    orders: [ [Promise], [Promise] ]
  },
  {
    customer_id: '123454351234123',
    name: 'Customer 2',
    orders: [ [Promise] ]
  }
]

我如何一次解决所有这 3 个 promise 并得到这个:

[
    {
        customer_id: '19847743234730384',
        name: 'Customer 1',
        orders: [
            {
                order_id: '98749873244324',
                price_per: 12,
                order_total: 34
            },
            {
                order_id: '9874987323545',
                price_per: 16,
                order_total: 34
            }
        ]
    },
    {
        customer_id: '123454351234123',
        name: 'Customer 2',
        orders: [
            {
                order_id: '918741433423',
                price_per: 6,
                order_total: 34
            }
        ]
    }
]

考虑到 Promise 不在一个数组中，我应该在哪里使用 Promise.all？

Answer 1

1. 将 customer_array.map 设为异步。
2. 将 await Promise.all 与 customer.orders.map 结合使用，以便返回结果而不是 Promise。
3. 由于我们已将 customer_array.map 设为 async，并且它没有包装在 async 函数中，因此我们需要解决 使用 Promise.all().then()。在 then 中写入所需的代码。

试试下面的方法。

const getOrderSum = order_id => {
  return new Promise(resolve => {
    setTimeout(() => resolve(34), 1000)
  })
}

const customer_array = [{
    customer_id: '19847743234730384',
    name: 'Customer 1',
    orders: [{
        order_id: '98749873244324',
        price_per: 12
      },
      {
        order_id: '9874987323545',
        price_per: 16
      }
    ]
  },
  {
    customer_id: '123454351234123',
    name: 'Customer 2',
    orders: [{
      order_id: '918741433423',
      price_per: 6
    }]
  }
]

const promises = customer_array.map(async customer => {
  customer.orders = await Promise.all(customer.orders.map(async order => {
    order.order_total = await getOrderSum();
    return order;
  }));
  return customer;
});

Promise.all(promises).then(results => console.log(results));