如何简化数组解构

问题描述

eslint持续显示prefer-restructuring错误。但是，我真的不知道数组解构如何工作，不希望有什么帮助。

这是两行返回错误：

word.results.inCategory = word.results.inCategory[0];

// and:

word.results = word.results.filter(
 (res) =>
  Object.keys(res).includes('partOfSpeech') &&
  Object.keys(res).includes('inCategory')
)[0];

再次，我在这方面不是很了解，所以将特别感谢您提供有关如何解决/简化此问题的帮助！

---

编辑：以下是示例对象供参考：

{
  word: 'midrash',results: [{
    deFinition: '(Judaism) an ancient commentary on part of the Hebrew scriptures that is based on Jewish methods of interpretation and attached to the biblical text',partOfSpeech: 'noun',inCategory: ['judaism'],typeOf: [ 'comment','commentary' ]
  },{ 
    deFinition: 'something',partOfSpeech: 'something',}],syllables: { count: 2,list: [ 'mid','rash' ] },pronunciation: { all: "'mɪdrɑʃ" },frequency: 1.82
}

解决方法

如果您已经确定自己的数据结构正确并且word.results.inCategory和word.results都是数组，那么您可以这样做：

const { results:{ inCategory: [inCategory] }} = word;
word.results.inCategory = inCategory;

// and:

const [results] = word.results.filter(
 (res) =>
  Object.keys(res).includes('partOfSpeech') &&
  Object.keys(res).includes('inCategory')
);

word.results = results;

当然，在第二次销毁过滤器时，您可以使用find来直接设置word.results而不进行销毁：

word.results = word.results.find(
 (res) =>
  Object.keys(res).includes('partOfSpeech') &&
  Object.keys(res).includes('inCategory')
);

,

要获取inCategory的值，您应按以下方式使用解构分配：

const obj = {
  word: 'midrash',results: {
    definition: '(Judaism) an ancient commentary on part of the Hebrew scriptures that is based on Jewish methods of interpretation and attached to the biblical text',partOfSpeech: 'noun',inCategory: 'judaism',typeOf: [ 'comment','commentary' ]
  },syllables: { count: 2,list: [ 'mid','rash' ] },pronunciation: { all: "'mɪdrɑʃ" },frequency: 1.82
}

let {results: {inCategory: category}} = obj;

//Now you can assign the category to word.results.inCategory
console.log(category);

对于过滤方法，我建议使用函数Array.prototype.find

word.results = word.results.find(
 (res) =>
  Object.keys(res).includes('partOfSpeech') &&
  Object.keys(res).includes('inCategory')
);