从数组中的嵌套对象中删除对象属性

Posted 2023-02-22

【中文标题】从数组中的嵌套对象中删除对象属性

【英文标题】：Remove object property from the nested object in the array

【发布时间】：2021-10-29 12:08:43

【问题描述】：

我有以下书籍。我想写一个函数来接收目标书和语言，从嵌套对象中删除该书（键，值）并返回新数组。

let books = [
  
    language: 'spanish',
    books: 
      book_1: 'book1_spanish',
      book_2: 'book2_spanish',
      book_3: 'book3_spanish'
    
  ,
  
    language: 'italian',
    books: 
      book_1: 'book1_italian',
      book_2: 'book2_italian',
      book_3: 'book3_italian'
    
  
];

let targetBook =  book_1: 'book1_spanish' ;
let language = 'spanish';

我被困在循环嵌套对象上。

function removeTargetBook(lan, target) 
  return books.map(book => 
    if (book.language == lan) 
      Object.values(book).map(value => );
    
  );

它应该返回没有 book_1 的相同数组：'book1_spanish'。 这是我的堆栈闪电战：https://stackblitz.com/edit/js-yxcrbq

谢谢。

【问题讨论】：

看看这个：developer.mozilla.org/en-US/docs/Web/javascript/Reference/…

知道了，不能做套料部分。

我知道了，看看答案

【参考方案1】：

您想使用delete 运算符。

我冒昧地扩展了您的代码，以向您展示如何在 forEach 循环中使用对象的 key 比进行键/值对比较更有效。另请注意，我将原始 books 对象保持不变，这对于单元可测试代码和数据完整性很重要，但您可以根据需要更改/实现。

let books = [
  
    language: 'spanish',
    books: 
      book_1: 'book1_spanish',
      book_2: 'book2_spanish',
      book_3: 'book3_spanish'
    
  ,
  
    language: 'italian',
    books: 
      book_1: 'book1_italian',
      book_2: 'book2_italian',
      book_3: 'book3_italian'
    
  
];

let targetBook =  book_1: 'book1_spanish' ;
let language = 'spanish';
let target = Object.keys(targetBook)[0]; // we only need the 'key' of the targetBook
let result;

function removeTargetBook(books, lan, target) 
  books.forEach(elem =>  // some instead of forEach works here as well
    elem.language === lan && (delete elem.books[target]);
  );
  
  return books;

// multiple tests for correctness
result = removeTargetBook(books, language, 'fool proof test');
console.log(result);
result = removeTargetBook(books, language, target);
console.log(result);
result = removeTargetBook(books, language, 'book_2');
console.log(result);
result = removeTargetBook(books, language, 'book_3');
console.log(result);

【讨论】：

谢谢，帮了大忙。

@zana10 有时您不需要函数，在上述解决方案中，除非您在多个地方“删除书籍”，否则我不会将函数用于单个回调

所使用的数据结构不允许单个项目搜索，因为所有内容都存储在对象数组中，我会将数据结构重新排列为一个大对象，键是语言和值作为书籍的内容，正如@RicardoSaracino 提到的那样，您将能够绕过回调函数的需要以及按语言匹配的循环

【参考方案2】：

我的版本太复杂了：

let books = [
    language: 'spanish',
    books: 
      book_1: 'book1_spanish',
      book_2: 'book2_spanish',
      book_3: 'book3_spanish'
    
  ,
  
    language: 'italian',
    books: 
      book_1: 'book1_italian',
      book_2: 'book2_italian',
      book_3: 'book3_italian'
    
  
];

let targetBook = 
  book_1: 'book1_spanish'
;
let language = 'spanish';

function removeTargetBook(lan, target) 
  return books.map(book => 
    if (book.language === lan) 

      return 
        language: book.language,
        books: Object.assign(...Object.keys(book.books).map((key) => (
          [key]: book.books[key]
        )).filter((e) => (Object.keys(e)[0] !== Object.keys(target)[0]) && Object.values(e)[0] !== Object.values(target)[0])),
      
     else 
      return book
    
  )

console.log(removeTargetBook(language, targetBook))

【参考方案3】：

我认为这样的事情应该可以工作

function removeTargetBook(lan, target) 
  books.some(book => 
    if (book.language == lan) 
      delete book.books[Object.keys(target)[0]];
      return true;
    
  );
  return books

【讨论】：

请添加更多详细信息以扩展您的答案，例如工作代码或文档引用。