Question

我知道下面的代码是不正确的，但是我认为它可以表明意图。这可能吗？我希望此函数更新原始数组，所有orderno的格式都为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

您要修改数组，因此需要

迭代数组的所有元素
获取元素的orderno属性的值
修改值
更新回数组元素的orderno属性

（1）遍历数组的所有元素

存在不同的可能性，最简单的技术是使用您已经尝试过的方法与for循环

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

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

 let orn = cars[i].orderno;

（3）修改其值

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

（4）在数组上直接分配新值

cars[i].orderno = orn;

因此，将所有部分放在一起

 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cars数组的元素并将逻辑上的orderno数字分成5组。

在regExp中，^表示字符串的开头，（）定义了一个逻辑组，以后可以将其用作变量，。{number}个字符，以及$字符串的末尾。因此，我们的正则表达式将一组3个字符分成$ 1，将一组1个字符分成$ 2，依此类推。然后，我们仅将初始字符串替换为“缝合”逻辑组。

RegExp乍一看似乎有些令人困惑，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，您可以使用map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}))

是否可以将特定数据添加到对象的属性？

4 个答案: