如何基于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"
    }
  ]
}]

为每个驾驶员将模型名称复制到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"
    }
  ]
}]

我的想法是遍历驾驶员，然后遍历每个驾驶员的汽车，然后遍历每辆汽车，然后遍...如此之多的循环。 我希望有一个更简单的方法来解决这个问题。

Answer 1

由于数据结构，您在这里无法避免循环。 但是你可以用一种很干净的方式

 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是品牌列表，而每个品牌都有型号列表？ 为什么cars数组看起来不像driver.cars数组，只是一个平面列表，每个对象都有Brand和Model？ 这将使搜索结构更加容易。 ModelID在各个品牌之间也应该是唯一的，因此您只需检查型号ID即可知道，如果型号相同，品牌也必须相同。

Answer 2

您可以使用两个forEach()循环来执行此操作，以循环驱动程序数组中的每个对象，然后循环当前对象内的cars数组。 之后，您可以使用两个find()循环在具有相同brand_id cars数组中查找对象，然后在内部模型中查找具有相同model_id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 3

您可以为汽车使用哈希表，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

另一个使用Array.prototype.reduce和哈希表的解决方案-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
)

用英语：

将驾驶员对象中的汽车列表中的每辆汽车的名称设置为以下结果：首先在具有正确品牌的汽车对象中查找汽车，然后在该汽车中找到与model_id匹配的模型，并取其名称。