按索引调整数组大小

Question

我有一个类似...的数组

let array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

...，一个起始索引 let start = 2 和一个结束索引 let end = 5。

我想调整以索引 startIndex 开始并以 endIndex 结束的数组的大小，如下例所示：

start = 2;
end = 3;

let result = [2, 3]

---

start = 2;
end = 2;

let result = [2]

---

start = 0;
end = 8;

let result = [0, 1, 2, 3, 4, 5, 6, 7, 8]

下面的这个 sn-p 是我到目前为止所得到的。但显然存在一些问题：

let array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

let start = 2
let end = 3
array.splice(0, start);
array.splice(end, array.length);

console.log(array)

Answer 1

您可以使用Array.slice(start,end) 并将end 加一，因为您希望它具有包容性。

let array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

let start = 2
let end = 3
var result = array.slice(start, end+1);

console.log(result)

Answer 2

您可以使用Array.slice() 获取新数组：

const array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

const start = 2
const end = 3

const result = array.slice(start, end + 1);

console.log(result)

如果你想直接改变原始数组，你可以将它与Array.splice()结合起来：

const array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

const start = 2
const end = 3

array.splice(0, array.length, ...array.slice(start, end + 1));

console.log(array)

Answer 3

当您将结果数组（第一次拼接后的数组）与实际数组 [有问题的数组] 的索引拼接时，您会得到该输出 请看下面的代码。运行这段代码，你就明白了

let array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

let start = 2
let end = 3

array.splice(0,start);

//After splice this is result array [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] 

console.log(array);

//If you splice again with end=3 index you get the result array as [2, 3, 4]. 
// To get your desired output as array you should work on array after you splice first time

array.splice(2,array.length);

console.log(array);

我希望这可以解释您的代码会发生什么。作为关于拼接方法的提醒。 splice() 方法在数组中添加/删除项目，并返回删除的项目。

语法：

array.splice(index, howmany, item1, ....., itemX)

Answer 4

您可以使用Array#copyWithin 并调整数组的length。

var start = 2,
    end = 5,
    array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13];

array.copyWithin(0, start, end + 1).length = end - start + 1;

console.log(array);