如何忽略数组解构的某些返回值? [英] How can I ignore certain returned values from array destructuring?

问题描述

当我只对索引 0 以外的数组值感兴趣时，是否可以避免在数组解构时声明无用的变量?

Can I avoid declaring a useless variable when array destructuring when I am only interested in array values beyond index 0?

在下面，我想避免声明 a，我只对索引 1 及以后感兴趣.

In the following, I want to avoid declaring a, I am only interested in index 1 and beyond.

// How can I avoid declaring "a"?
const [a, b, ...rest] = [1, 2, 3, 4, 5];

console.log(a, b, rest);

推荐答案

当我只对索引 0 以外的数组值感兴趣时，是否可以避免在数组解构时声明无用的变量?

Can I avoid declaring a useless variable when array destructuring when I am only interested in array values beyond index 0?

是的，如果您将作业的第一个索引留空，则不会分配任何内容.此行为在此处解释.

Yes, if you leave the first index of your assignment empty, nothing will be assigned. This behavior is explained here.

// The first value in array will not be assigned
const [, b, ...rest] = [1, 2, 3, 4, 5];

console.log(b, rest);

您可以在任何地方使用任意数量的逗号，除了在休息元素之后:

You can use as many commas as you like wherever you like, except after a rest element:

const [, , three] = [1, 2, 3, 4, 5];
console.log(three);

const [, two, , four] = [1, 2, 3, 4, 5];
console.log(two, four);

以下产生错误:

const [, ...rest,] = [1, 2, 3, 4, 5];
console.log(rest);

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