如何根据引用的值从另一个对象扩展一个对象？

Question

我有两个数组，每个元素都是一个对象：

var cars = [{
  brand: "VW",
  brand_id: "1",
  models: [
    {
      name: "Golf",
      model_id: "1"
    },
    {
      name: "Passat",
      model_id: "2"
    }
  ]
},
{
  brand: "BMW",
  brand_id: "2",
  models: [
    {
      name: "X3",
      model_id: "1"
    },
    {
      name: "X5",
      model_id: "2"
    }
  ]
}];

var drivers = [{
  name: "Test Driver",
  cars: [
    {
      brand_id: "1",
      model_id: "1"
    },
    {
      brand_id: "2",
      model_id: "1"
    }
  ]
}]

将模型名称复制到每个驾驶员的汽车对象中的最佳方法是什么？这是预期的结果：

var drivers = [{
  name: "Test",
  cars: [
    {
      brand_id: "1",
      model_id: "1",
      name: "Golf"
    },
    {
      brand_id: "2",
      model_id: "1",
      name: "X3"
    }
  ]
}]

我的想法是循环遍历驾驶员，然后循环遍历每个驾驶员的汽车，然后循环遍历每辆汽车，然后循环......这么多循环。我希望有更简单的方法来解决这个问题。

Answer 1

您可以使用两个forEach()loop 来执行此操作，循环驱动程序数组中的每个对象，然后循环当前对象内的汽车数组。之后，您可以使用两个find() 循环在汽车数组中查找具有相同brand_id 的对象，然后在模型内部查找具有相同model_id 的对象并返回name。

var cars = [{"brand":"VW","brand_id":"1","models":[{"name":"Golf","model_id":"1"},{"name":"Passat","model_id":"2"}]},{"brand":"BMW","brand_id":"2","models":[{"name":"X3","model_id":"1"},{"name":"X5","model_id":"2"}]}];
var drivers = [{"name":"Test Driver","cars":[{"brand_id":"1","model_id":"1"},{"brand_id":"2","model_id":"1"}]}];

drivers.forEach(function(e) {
  e.cars.forEach(function(c) {
    c.name = cars.find(function(a) {
      return a.brand_id == c.brand_id;
    }).models.find(function(a) {
      return c.model_id == a.model_id;
    }).name
  })
})

console.log(drivers)

Answer 2

由于数据结构，您无法避免此处出现循环。但是你可以用一种相当干净的方式来做到这一点

var cars = [{"brand":"VW","brand_id":"1","models":[{"name":"Golf","model_id":"1"},{"name":"Passat","model_id":"2"}]},{"brand":"BMW","brand_id":"2","models":[{"name":"X3","model_id":"1"},{"name":"X5","model_id":"2"}]}];
var drivers = [{"name":"Test Driver","cars":[{"brand_id":"1","model_id":"1"},{"brand_id":"2","model_id":"1"}]}];

drivers.forEach(expandCars);

function expandCars(driver) {
  // map each car to list of car brands filtered by brand ID
  // then to resulting list of car models by model ID
  driver.cars = driver.cars.map(car => {
     var carModels = cars.filter(carBrand => carBrand.brand_id == car.brand_id)[0].models;
     var model = carModels.filter(carModel => carModel.model_id == car.model_id)[0];
    
     return Object.assign(car, model);
  });
}



console.log(drivers);

老实说，这种数据结构看起来不必要地复杂。为什么cars是一个Brands列表，而每个Brand都有一个Models列表？为什么不让cars 数组看起来像driver.cars 数组，只是一个每个对象都有品牌和型号的平面列表？这将使搜索汽车的结构变得更加容易。 ModelID在品牌之间也应该是唯一的，所以你只需检查型号ID就知道如果型号相同则品牌必须相同。

Answer 3

您可以对汽车使用哈希表，然后再分配名称。

var cars = [{ brand: "VW", brand_id: "1", models: [{ name: "Golf", model_id: "1" }, { name: "Passat", model_id: "2" }] }, { brand: "BMW", brand_id: "2", models: [{ name: "X3", model_id: "1" }, { name: "X5", model_id: "2" }] }],
    drivers = [{ name: "Test Driver", cars: [{ brand_id: "1", model_id: "1" }, { brand_id: "2", model_id: "1" }] }],
    hash = Object.create(null);

cars.forEach(function (brands) {
    brands.models.forEach(function (model) {
        hash[[brands.brand_id, model.model_id].join('|')] = model;
    });
});

drivers.forEach(function (driver) {
    driver.cars.forEach(function (car) {
        car.name = hash[[car.brand_id, car.model_id].join('|')].name;
    });
});

console.log(drivers);

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Answer 4

使用Array.prototype.reduce 和哈希表的另一种解决方案 - 参见下面的演示：

var cars=[{brand:"VW",brand_id:"1",models:[{name:"Golf",model_id:"1"},{name:"Passat",model_id:"2"}]},{brand:"BMW",brand_id:"2",models:[{name:"X3",model_id:"1"},{name:"X5",model_id:"2"}]}];
var drivers=[{name:"Test Driver",cars:[{brand_id:"1",model_id:"1"},{brand_id:"2",model_id:"1"}]}];

var result = drivers.reduce(function(hash) {

  // create a hash table first
  cars.forEach(function(element) {
    element.models.forEach(function(model) {
      hash[element.brand_id + "|" + model.model_id] = model.name;
    });
  });

  // now reduce it to obtain the result
  return function(prev, curr) {
    curr.cars.forEach(function(element) {
      element.name = hash[element.brand_id + "|" + element.model_id];
    });
    prev.push(curr);
    return prev;
  };
}(Object.create(null)), []);

console.log(result);

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Answer 5

drivers.cars.forEach(                                
  car =>                                             
    car.name =                                       
      cars.find(                                     
        car2 =>                                      
          car2.brand_id === car.brand_id             
      )
        .models.find(                                
          model => 
            model.model_id === car.model_id
        ).name
)

英文：

将drivers对象的cars列表中的每辆车的名称设置为首先在cars对象中查找具有正确品牌的汽车，然后在该汽车中找到model_id匹配的模型，并取它的名字。