Question

let traveler = [{
    timestamp: 'qualia',
    firstname: 'unborn',
    lastname: 'child',
    location: 'null'
  },
  {
    timestamp: 1000,
    firstname: 'Olivia',
    lastname: 'Kirshner',
    location: 'Titan'
  },
  {
    timestamp: 1001,
    firstname: 'James',
    lastname: 'Cole',
    location: 'Emerson Hotel'
  }
]

// function noerror(splinter = traveler[0]) {
//     return [{firstname, lastname}] = splinter
// }

function restructure(time) {
  // if no array is returned on the lookup, and the value flags as undefined in restructure()
  let missingLink = traveler[0]
  let tempArray = traveler
    .filter((item) => item.timestamp === time)
  try {
    if (tempArray.length === 0) {
      return {
        firstname,
        lastname
      } = missingLink
    }
    return [{
      firstname,
      lastname
    }] = tempArray
  } catch (e) {
    // e instanceof TypeError // boolean error type check
    switch (e.name) {
      case 'TypeError':
        console.error(`Could not complete your request: ${e.message}`);
        return {
          firstname,
          lastname
        } = missingLink
      default:
        break;
    }
  }
}

// the logging statement breaks the code with a ReferenceError: 'timestamp' is not defined
// does that mean destructuring enforces type safety???
// console.log(traveler)
[{timestamp, firstname, lastname, location}] = traveler

restructure(1001)
console.log(`firstname: ${firstname}, lastname: ${lastname}`)

restructure(100)
console.log(`firstname: ${firstname}, lastname: ${lastname}`)

restructure(1000)
lastname = 'Redforester' // temp assignment
console.log(`firstname: ${firstname}, lastname: ${lastname}`)

restructure(1000)

// console.log(traveler)
[{timestamp, firstname, lastname, location}] = traveler

console.log(`timestamp ${timestamp}, firstname ${firstname}, lastname ${lastname}, location ${location}`);

解构强制类型安全性是否解释了为什么如果包含console.log(traveler)时此代码不起作用？

您仍可以在发生解构的对traveler的函数调用之后记录对象restructure()，但除非在进行任何解构操作之前完成对变量restructure()的解构分配，在旅行者对象上，它将出错。换句话说，必须在对对象进行任何其他解构之前，timestamp才能显示：[{timestamp, firstname, lastname, location}] = traveler。

将日志记录语句保留在第70行的结果如下：

发生异常：ReferenceError

ReferenceError：未在对象上定义时间戳。

不幸的是，在这里的代码控制台中，结果没有像在vscode中一样显示出来。

最终，我正在寻找一种在一段时间后引用traveler中的每个键的方法，而不必将所有变量解构为在其他解构方案中使用该对象的前提。

我也对与将键/值对动态添加到同一对象中的对象感到好奇。例如，如果类型不匹配其他“记录”，以后添加键/值对是否会破坏结构？

销毁强制类型安全说明代码无法按预期工作吗？

0 个答案: