ES6解构：如何创建一个省略动态引用键的新对象

Question

当关键引用是动态的时，是否存在使用解构和扩展运算符的ES6（和向上）解决方案来创建具有从原始对象删除的键和值的新对象，因此：

const state = {
   12344: {
      url: 'http://some-url.com',
      id: '12344'
   },
   12345: {
      url: 'http://some-other-url.com',
      id: '12345'
   }
}

const idToDelete = 12344

const { [idToDelete], ...newState } = state // dynamic key

console.log('newState:', newState)

// desired newState would only have the key 12345 and its value

除非是我现在的Babel设置，否则我无法弄清楚干净的ES6方式（如果存在）。

提前谢谢了

Answer 1

当使用动态id进行解构时，您需要使用remove值设置var：the doc about this

const state = {
   12344: {
      url: 'http://some-url.com',
      id: '12344'
   },
   12345: {
      url: 'http://some-other-url.com',
      id: '12345'
   }
}

const idToDelete = 12344

// the removed object will go to unusedVar
const { [idToDelete]: unusedVar, ...newState } = state // dynamic key

console.log('newState:', newState)

如果您不需要保留已删除的对象，则更好的方法是使用关键字delete

const state = {
   12344: {
      url: 'http://some-url.com',
      id: '12344'
   },
   12345: {
      url: 'http://some-other-url.com',
      id: '12345'
   }
}

const idToDelete = 12344

delete state[idToDelete]

console.log('newState:', state)

Answer 2

我不认为用ES6解构可以干净利落。由于其他答案包括改变状态，请尝试以下方法：

const state = {
   12344: {
      url: 'http://some-url.com',
      id: '12344'
   },
   12345: {
      url: 'http://some-other-url.com',
      id: '12345'
   }
}

const idToDelete = 12344

const newState = Object.assign({}, state);
delete newState[idToDelete];

console.log('newState:', newState)
console.log('old state:', state);