从数组中删除 object 并同时更改其属性之一 - javascript

Question

我有这个对象数组：

const flights = [
    { id: 00, to: "New York", from: "Barcelona", cost: 700, scale: false },
    { id: 01, to: "Los Angeles", from: "Madrid", cost: 1100, scale: true },
    { id: 02, to: "Paris", from: "Barcelona", cost: 210, scale: false },
    { id: 03, to: "Roma", from: "Barcelona", cost: 150, scale: false },
    { id: 04, to: "London", from: "Madrid", cost: 200, scale: false },
    { id: 05, to: "Madrid", from: "Barcelona", cost: 90, scale: false },
    { id: 06, to: "Tokyo", from: "Madrid", cost: 1500, scale: true },
    { id: 07, to: "Shangai", from: "Barcelona", cost: 800, scale: true },
    { id: 08, to: "Sydney", from: "Barcelona", cost: 150, scale: true },
    { id: 09, to: "Tel-Aviv", from: "Madrid", cost: 150, scale: false },
  ];

我想要做的是，当我delete其中一个对象时，对于那些在删除 object 之后出现的对象，将它们的id减 1。

例如，如果我要delete第五个 object ( id: 04 )，我希望 ids 从 0 到 8 (0, 1, 2, 3, 4, 5, 6, 7, 8) 而不是 go从 0 到 9 并跳过 4（0、1、2、3、5、6、7、8、9）。

这是我目前的代码：

let flightIdToDelete = 04;

for (let i = 0; i < flights.length; i++) {
    if (flightIdToDelete === flights[i].id) {
        delete flights[i];
    }
}

/* The array would then become:

const flights = [
    { id: 00, to: "New York", from: "Barcelona", cost: 700, scale: false },
    { id: 01, to: "Los Angeles", from: "Madrid", cost: 1100, scale: true },
    { id: 02, to: "Paris", from: "Barcelona", cost: 210, scale: false },
    { id: 03, to: "Roma", from: "Barcelona", cost: 150, scale: false },
    { id: 05, to: "Madrid", from: "Barcelona", cost: 90, scale: false },
    { id: 06, to: "Tokyo", from: "Madrid", cost: 1500, scale: true },
    { id: 07, to: "Shangai", from: "Barcelona", cost: 800, scale: true },
    { id: 08, to: "Sydney", from: "Barcelona", cost: 150, scale: true },
    { id: 09, to: "Tel-Aviv", from: "Madrid", cost: 150, scale: false },
  ]; */

我最好的猜测是从删除的对象开始遍历对象的 ID，然后减去 1，但我不太确定该怎么做。

请帮助！！！

Answer 1

你可以只使用一个filter ，这将是一个实际的同步变化：

const flights = [
    {id: 0, to: "New York", from: "Barcelona", cost: 700, scale: false},
    {id: 1, to: "Los Angeles", from: "Madrid", cost: 1100, scale: true},
    {id: 2, to: "Paris", from: "Barcelona", cost: 210, scale: false},
    {id: 3, to: "Roma", from: "Barcelona", cost: 150, scale: false},
    {id: 4, to: "London", from: "Madrid", cost: 200, scale: false},
    {id: 5, to: "Madrid", from: "Barcelona", cost: 90, scale: false},
    {id: 6, to: "Tokyo", from: "Madrid", cost: 1500, scale: true},
    {id: 7, to: "Shangai", from: "Barcelona", cost: 800, scale: true},
    {id: 8, to: "Sydney", from: "Barcelona", cost: 150, scale: true},
    {id: 9, to: "Tel-Aviv", from: "Madrid", cost: 150, scale: false},
];

const id = 4;
let count = 0;

const r = flights.filter((a, idx) => {
    const d = a.id !== id;
    count += d ? 0 : 1;
    a.id -= count;
    return d;
});

console.log(r);

 const flights = [ {id: 0, to: "New York", from: "Barcelona", cost: 700, scale: false}, {id: 1, to: "Los Angeles", from: "Madrid", cost: 1100, scale: true}, {id: 2, to: "Paris", from: "Barcelona", cost: 210, scale: false}, {id: 3, to: "Roma", from: "Barcelona", cost: 150, scale: false}, {id: 4, to: "London", from: "Madrid", cost: 200, scale: false}, {id: 5, to: "Madrid", from: "Barcelona", cost: 90, scale: false}, {id: 6, to: "Tokyo", from: "Madrid", cost: 1500, scale: true}, {id: 7, to: "Shangai", from: "Barcelona", cost: 800, scale: true}, {id: 8, to: "Sydney", from: "Barcelona", cost: 150, scale: true}, {id: 9, to: "Tel-Aviv", from: "Madrid", cost: 150, scale: false}, ]; const id = 4; let count = 0; const r = flights.filter((a, idx) => { const d = a.id;== id? count += d: 0; 1. a;id -= count; return d; }). console;log(r);

Answer 2

您可以在Array#filter方法和Array#map的帮助下简单地实现此要求。

现场演示：

 let flights = [ { id: 0, to: "New York", from: "Barcelona", cost: 700, scale: false }, { id: 1, to: "Los Angeles", from: "Madrid", cost: 1100, scale: true }, { id: 2, to: "Paris", from: "Barcelona", cost: 210, scale: false }, { id: 3, to: "Roma", from: "Barcelona", cost: 150, scale: false }, { id: 4, to: "London", from: "Madrid", cost: 200, scale: false }, { id: 5, to: "Madrid", from: "Barcelona", cost: 90, scale: false }, { id: 6, to: "Tokyo", from: "Madrid", cost: 1500, scale: true }, { id: 7, to: "Shangai", from: "Barcelona", cost: 800, scale: true }, { id: 8, to: "Sydney", from: "Barcelona", cost: 150, scale: true }, { id: 9, to: "Tel-Aviv", from: "Madrid", cost: 150, scale: false } ]; flights = flights.filter(({ id }) => id.== 4),map((obj. index) => { obj;id = index; return obj; }). console.log(flights)