重用通过解构创建的对象文字

Question

我正在尝试为两个异步调用重用对象文字。最后，我的期望应该检查 deleteBucket 调用是否成功。问题是我不能这样做，或者它说我已经定义了 dup 变量：

it('can delete a bucket', async () => {
      const options = { branch: '11' }

      let { failure, success, payload } = await deployApi.createBucket(options)
      let { failure, success, payload} = await deployApi.deleteBucket(options.branch)

      expect(success).to.be.true
    })

有人告诉我我可以在第二个周围放一个 () ，但这会导致错误TypeError: (0 , _context4.t0) is not a function：

it('can delete a bucket', async () => {
  const options = { branch: '11' }

  let { failure, success, payload } = await deployApi.createBucket(options)

  ({ failure, success, payload} = await deployApi.deleteBucket(options.branch))

  expect(success).to.be.true
})

这确实有效，但需要我更改我不想做的已解析对象的名称：

it('can delete a bucket', async () => {
      const options = { branch: '11' }

      let { failure, success, payload } = await deployApi.createBucket(options)
      let { failure1, success1, payload1} = await deployApi.deleteBucket(options.branch)

      expect(success1).to.be.true
    })

更新：

有人建议我在 const 行后面需要一个分号。没有什么区别，当我运行它时仍然遇到同样的错误：

Answer 1

缺少分号

你有两个(...)一起测序 -

await deployApi.createBucket(options)  

({ failure, success, payload} = await deployApi.deleteBucket(options.branch))

ES 解释器将其视为 -

await deployApi.createBucket(options)(... await deployApi.deleteBucket(options.branch))

这相当于 -

const r1 = deployApi.createBucket(options)
const r2 = r1(... await deployApi.deleteBucket(options.branch))
await r2

这与你的实际意图有很大不同 -

const r1 = await deployApi.createBucket(...)
const r2 = await deployApi.deleteBucket(...)

要重用let解构的对象，需要括号 -

// initial assignment
let { a, b, c } = ...

// without the parentheses, ES interprets as illegal assignment using =
{ a, b, c } = ...

// with parentheses, ES interprets as destructuring assignment
({ a, b, c } = ...)

如果重复使用相同的let绑定，则在不使用分号时，所需的括号会更改程序的含义。

it('can delete a bucket', async () => {
  const options = { branch: '11' }
                                                           // semicolon here
  let { failure, success, payload } = await deployApi.createBucket(options); 
  
  ({ failure, success, payload} = await deployApi.deleteBucket(options.branch))
  
  expect(success).to.be.true
})