使用对象数组解构对象

Question

我有这样的对象：

obj: {
  child1: [
    { type, checked, text, ... },
    { type, checked, text, ... },
    { type, checked, text, ... },
  ],
  child2: [
    { type, checked, text, ... },
    ...
  ],
  ...
}

我需要几乎相同的对象，但子元素应该具有仅由类型和检查值组成的对象。需要我的输出像下面的例子。

输出：

obj: {
  child1: [
    {
      type: "type",
      checked: "checked"
    },
    {
      type: "type",
      checked: "checked"
    },
    {
      type: "type",
      checked: "checked"
    }
  ],
  child2: [
    {
      type: "type",
      checked: "checked"
    }
  ]
}

到目前为止，我尝试过的所有方法似乎都不起作用。

我最后一次失败的尝试：

    Object.keys(tabs).forEach(key =>
      ({
        updatedState: {
          [key]: (({ documentTypeId, checked }) => ({ documentTypeId, checked }))(tabs[key]),
        },
      }),
    );

Answer 1

您可以使用 Array.reduce() to iterate the object's keys, with an inner Array.map() and destructuring 从您要保留的属性创建新对象：

const type = 'type'
const checked = 'checked'
const text = 'text'

const obj = {
  child1: [
    { type, checked, text },
    { type, checked, text },
    { type, checked, text },
  ],
  child2: [
    { type, checked, text },
  ],
}

const result = Object.keys(obj).reduce((r, k) => {
  r[k] = obj[k].map(({ type, checked }) => ({ type, checked }))
  
  return r
}, {})

console.log(result)

Answer 2

您可以结合使用 reduce（遍历 object 键）和 map （对于您的 children 阵列）

const obj = {
  child1: [
    { type: 1, checked: true, text: 'aaa'},
    { type: 2, checked: false, text: 'bbb'},
    { type: 3, checked: true, text: 'ccc'}
  ],
  child2: [
    { type: 4, checked: true, text: 'ddd'},
    { type: 5, checked: false, text: 'eee'},
    { type: 6, checked: true, text: 'fff'}
  ]
};

const result = Object.keys(obj).reduce((acc, key) => { 
  acc[key] = obj[key].map(child => 
    ({type: child.type, checked: child.checked}));
    
  return acc;
}, {}); 

console.log(result);

使用对象数组解构对象

Desctructuring object with object arrays

javascript

destructuring