Question

我最近开始尝试函数式编程，并且我正在尝试转换我使用命令式编程编写的旧模块。

假设我有两个对象数组，即

orders: [
  {
   idOrder: 1,
   amount: 100,
   customerId: 25,
  },
  {
   idOrder: 2,
   amount: 200,
   customerId: 20,
  }
]


customers: [
  {
   customerId: 20,
   name: "John Doe",
   orders: []
  },
  {
   customerId: 25,
   name: "Mary Jane",
   orders: []
  }
]

我想将所有订单推送给各自的客户。这样做有干净的方法吗？

我试过这个，但显然它不会这样：

customers.orders = orders.filter((x) => {
  if (x.customerId === customerId) {
    customer.orders.push(x);
  }
});

由于

Answer 1

您可以使用Map并先获取所有客户，然后将订单推送给客户。

var object = { orders: [{ idOrder: 1, amount: 100, customerId: 25 }, { idOrder: 2, amount: 200, customerId: 20 }], customers: [{ customerId: 20, name: "John Doe", orders: [] }, { customerId: 25, name: "Mary Jane", orders: [] }] },
    map = object.customers.reduce((m, a) => m.set(a.customerId, a), new Map);

object.orders.forEach(a => map.get(a.customerId).orders.push(a));

console.log(object.customers);

Answer 2

可能的解决方案：

for (c of customers){
    c.orders.push(orders.filter( function(o){ return o.customerId === c.customerId} ));

}

Answer 3

如果您认为customers为累加器，则Reduce orders可以customers作为初始值。

注意：如果您不希望将此作为克隆 customers的副作用，则会发生变异customers。还没有找到customerId的错误处理。

var orders =  [{ idOrder: 1, amount: 100, customerId: 25 }, { idOrder: 2, amount: 200, customerId: 20}];
var customers =  [{ customerId: 20, name: "John Doe", orders: [] }, { customerId: 25, name: "Mary Jane", orders: [] } ];

var customers_orders = orders.reduce( 
               (accum, v) => 
                   { accum.find(
                            c => c.customerId == v.customerId).orders.push(v);
                     return accum; 
                   }, customers);

console.log(customers_orders);

Answer 4

您可以编写一个函数并将其传递给reduce方法并使用map

组合它

只有一件事：一旦创建，它可能永远不会改变。您可以使用Object.assign和concat。

var customersWithOrders = customers.map(function(customer) {
  var relatedOrders = orders.filter(function(order) { return order.customerId === customer.customerId })
  return Object.assign(
    customer, 
    {
      orders: customer.orders.concat(relatedOrders)
    }
  )
})

Javascript函数式编程/传递参数

4 个答案: