我可以将特定数据添加到对象的属性吗？

Question

我知道下面的代码不正确，但我认为它会表明意图。这可能吗？我希望这个函数更新原始数组，所有 orderno 的格式都是 000.0.000.00000.0。

let cars= [ 
  {orderno: "5766302385925", make: "Alfa", dealership: "NA"},
  {orderno: "4663873261390", make: "Merc", dealership: "NA"},
  {orderno: "8458821291579" , make: "BMW", dealership: "EU"},
  {orderno: "3376768687480", make: "Ford", dealership: "NA"},
  {orderno: "5186132840921", make: "Buick", dealership: "EU"},
];

function addOrderDot() {
    for (var i=0; i<cars.length; i++) {
    cars.orderno[i].splice(2,0,'.')
    cars.orderno[i].splice(3,0,'.')
    cars.orderno[i].splice(6,0,'.')
    cars.orderno[i].splice(11,0,'.')
    } 
}

函数后所需的数组：

let cars= [ 
  {orderno: "576.6.302.38592.5", make: "Alfa", dealership: "NA"},
  {orderno: "466.3.873.26139.0", make: "Merc", dealership: "NA"},
  {orderno: "845.8.821.29157.9" , make: "BMW", dealership: "EU"},
  {orderno: "337.6.768.68748.0", make: "Ford", dealership: "NA"},
  {orderno: "518.6.132.84092.1", make: "Buick", dealership: "EU"},
];

Answer 1

一种方法是使用map() 和slice()。

let cars= [ 
  {orderno: "5766302385925", make: "Alfa", dealership: "NA"},
  {orderno: "4663873261390", make: "Merc", dealership: "NA"},
  {orderno: "8458821291579", make: "BMW", dealership: "EU"},
  {orderno: "3376768687480", make: "Ford", dealership: "NA"},
  {orderno: "5186132840921", make: "Buick", dealership: "EU"},
];

function addOrderDot(cars) {
  return cars.map(car => ({
    ...car,
    orderno: car.orderno.slice(0, 3) + '.' +
      car.orderno.slice(3, 4) + '.' +
      car.orderno.slice(4, 7) + '.' +
      car.orderno.slice(7, 12) + '.' +
      car.orderno.slice(12, 13)
  }));
}

let result = addOrderDot(cars);

console.log(result);

Answer 2

你想就地修改你的数组，所以你需要

1. 遍历数组的所有元素
2. 取元素的orderno属性的值
3. 修改数值
4. 更新回数组元素的 orderno 属性

---

(1) 遍历数组的所有元素

有不同的可能性，最简单的技术是循环使用for，就像你已经尝试过的一样

 for (let i=0; i<cars.length; i++) {
 } 

(2) 使用cars[i] 寻址数组元素，并使用cars[i].orderno 访问errorno 属性

 let orn = cars[i].orderno;

(3)修改其值

orn = orn.slice(0,3)+'.'+orn.slice(3,4);

(4) 直接给数组赋值

cars[i].orderno = orn;

所以，把所有部分放在一起

 for (var i=0; i<cars.length; i++) {
    let orn = cars[i].orderno;
    orn = orn.slice(0,2)+'.'+orn.slice(2,4) ;
    cars[i].orderno = orn;
 }

Answer 3

更优雅的解决方案是使用简单的正则表达式来替换每个字符串：

let cars= [ 
  {orderno: "5766302385925", make: "Alfa", dealership: "NA"},
  {orderno: "4663873261390", make: "Merc", dealership: "NA"},
  {orderno: "8458821291579" , make: "BMW", dealership: "EU"},
  {orderno: "3376768687480", make: "Ford", dealership: "NA"},
  {orderno: "5186132840921", make: "Buick", dealership: "EU"},
];

function addOrderDot() {
    cars.forEach(car => {
       car.orderno = car.orderno.replace(
      /^(.{3})(.{1})(.{3})(.{5})(.{1})$/,
      `$1.$2.$3.$4.$5`
    )
    })
}

说明：我们遍历汽车数组的元素，并将 orderno 数字逻辑分组为 5 组。

在正则表达式中，^ 表示字符串的开头，() a 定义了我们以后可以用作变量的逻辑组，.{number} 表示字符数，$ 表示字符串的结尾。 所以我们的正则表达式将一组 3 个字符分隔为 $1，一组 1 个字符分隔为 $2，依此类推。 然后我们只需将初始字符串替换为“缝合”的逻辑组。

RegExp 起初可能看起来有点混乱，但在处理字符串和明显的模式时可能是一个很好的工具。

Answer 4

首先你可以使用map来转换数组：

let cars= [ 
  {orderno: "5766302385925", make: "Alfa", dealership: "NA"},
  {orderno: "4663873261390", make: "Merc", dealership: "NA"},
  {orderno: "8458821291579" , make: "BMW", dealership: "EU"},
  {orderno: "3376768687480", make: "Ford", dealership: "NA"},
  {orderno: "5186132840921", make: "Buick", dealership: "EU"},
];

console.log(cars.map(car => {car.orderno = "test"; return car}))