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

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问题描述

当我只对索引0之外的数组值感兴趣时,我可以避免在数组解构时声明无用变量吗?



在下面,我想避免声明 a ,我只对索引1及以后感兴趣。



  //如何避免声明a?const [a,b,... rest] = [1,2,3,4,5]; console.log(a, b,休息);  

解决方案

< blockquote>

当我只对索引0之外的数组值感兴趣时,我可以避免在数组解构时声明无用变量吗?


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。此行为是此处说明的



  //数组中的第一个值不会被赋值为[,b ,. ..rest] = [1,2,3,4,5]; console.log(b,rest);  



除了休息元素之外,您可以随意使用任意数量的逗号:



  const [,,三] = [1,2,3,4,5]; console.log(三); const [,two,,four ] = [1,2,3,4,5]; console.log(二,四);  



T.他跟随产生错误:



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


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

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);

解决方案

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

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);

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);

The following produces an error:

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

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