Question

以下代码以两种方式声明了字符串的Flowtype Union。 1）使用内置的Union Type，需要再次输入所有代码的警告，创建一个重复。 2）利用$Keys：Flow v0.38.0正确推断 makeObjectWithKeys 类型，但我想知道是否可以手动编写这样的注释。

const codesArray = [
   {
     name: 'Lorem',
     code: 'lm'
   },
   {
     name: 'Ipsum',
     code: 'ip'
   },
   // ...
 ]

// Define the CodeType "manually" with the Union built-in
type CodeTypeManual =
   | "lm"
   | "ip"
   // ...

const noErrorManual: CodeTypeManual = 'lm'
const flowErrorPropertyNotFoundManual: CodeTypeManual = 'zz'

// Define the CodeType by taking advantage of $Keys
const makeObjectWithKeys = (inArray) => { // Type annotations?
  return inArray.reduce(
        (objAcc, curObj) => { // Type annotations?
          const retObj = { ...objAcc }
          const { code } = curObj
          retObj[code] = code
          return retObj
        }
        , {}
  )
}
const objectWithCodesAsKeys = makeObjectWithKeys(codesArray)

type CodeType = $Keys<typeof objectWithCodesAsKeys>

let noError: CodeType = 'ip'
let flowErrorPropertyNotFound: CodeType = 'zz'

Answer 1

如果我理解您的问题，则以下内容可能有所帮助。

/* @flow */
const codesArray = [
   {
     name: 'Lorem',
     code: 'lm'
   },
   {
     name: 'Ipsum',
     code: 'ip'
   },
   // ...
 ]

const makeObjectWithKeys = function<T: Object>(inArray: Array<T>, key: $Keys<T>) {
  return inArray.reduce((objAcc: { [name: typeof key]: T }, curObj: T) => {
    return Object.assign({}, objAcc, { [curObj[key]]: curObj })
  }, {})
}

const objectWithCodesAsKeys = makeObjectWithKeys(codesArray, 'code')

我可以为此函数编写Flow类型注释，将对象数组减少到对象

1 个答案: