Question

我想将初始数据转换为工作数据。两者都有自己的类型，唯一的区别是在初始数据中，名称是可选的。当我创建工作数据时，我使用默认值'__unknown__'作为空名称。

以下是示例代码：

/* @flow */

type NAME_OPTIONAL = {
  name?: string
}

type NAME_MANDATORY = {
  name: string
}

type INITIAL = {
  source: string,
  data: NAME_OPTIONAL[] // <-- Here the names are OPTIONAL.
}

type WORKING = {
  source: string,
  data: NAME_MANDATORY[] // <-- Here the name are MANDATORY.
}

// We have some initial data.
const initial: INITIAL = {
  source: 'some.server.com',
  data: [{ name: 'Adam' }, { name: undefined }]
}

// And we want to turn initial data into working data.
const workingData = initial.data.map((i) => {
  return { 
    name: i.name || '__unknown__'
  }
});

// This is OK:
const working1: WORKING = {
  source: initial.source, 
  data: workingData
}

// This is NOT OK:
const working2: WORKING = {
  ...initial,
  data: workingData
}

在上面的示例结尾处初始化working1是正常的，但使用对象扩展运算符初始化working2会导致flowtype显示此错误：

4:  name?: string
            ^ undefined. This type is incompatible with
8:  name: string
           ^ string

我不明白传播运营商如何导致这种情况。任何人都可以解释一下吗？谢谢。

"Working" example on https://flowtype.org/try/... is here

Answer 1

有a lot of bugs about the spread operator。您的案例似乎与this one相同。

除非用…替换Object.assign运算符，否则在修复它之前可能没有解决方案：

const working2: WORKING = Object.assign({}, initial, { data: workingData })

如果仍然无效，您可以在该行上方添加注释：

// $FlowIssue
const working2: WORKING = Object.assign({}, initial, { data: workingData })

或者：

// $FlowIssue
const working2: WORKING = {
  ...initial,
  data: workingData
}

然后在.flowconfig：

中添加此设置

[options]
suppress_comment=.*\\$FlowIssue

这将抑制错误。

在具有可选属性的对象中使用spread运算符时出错

1 个答案: