IT技术 - Typescript React：props的联合类型在提供多余的属性时不会显示错误 - 吾爱随笔录

我正在尝试使用联合类型作为 React 组件的props

type Props =
  | {
      type: "string"
      value: string
    }
  | {
      type: "number"
      value: number
    }
  | {
      type: "none"
    }

class DynamicProps extends React.Component<Props> {
  render() {
    return null
  }
}

// Ok
const string_jsx = <DynamicProps type="string" value="hello" />

// Error as expected, value should be a string
const string_jsx_bad = <DynamicProps type="string" value={5} />

// Ok
const number_jsx = <DynamicProps type="number" value={5} />

// Error as expcted value should be a number
const number_jsx_bad = <DynamicProps type="number" value="hello" />

// Error as expected, invalid isn't a property on any of the unioned types
const extra = <DynamicProps type="string" value="extra" invalid="what" />

// No error? There should be no value when type="none"
const none_jsx = <DynamicProps type="none" value="This should be an error?" />

// Ok, seems like value has become optional
const none2_jsx = <DynamicProps type="none" />

// Error as expected, value is not present. Value doesn't seem to be made optional all the time
const required = <DynamicProps type="string" />

它似乎部分起作用，因为根据typeprops有效props会发生变化。然而，虽然没有出现在联合中的任何类型上的额外属性将是一个错误，但似乎出现在至少一个联合类型中但不属于基于判别属性的类型不会是错误。

我不确定为什么会这样。使用联合类型作为react组件的props是一种反模式吗？

type Props = | { type: "string" value: string } | { type: "number" value: number } | { type: "none" } type UnionKeys<T> = T extends any ? keyof T : never; type StrictUnionHelper<T, TAll> = T extends any ? T & Partial<Record<Exclude<UnionKeys<TAll>, keyof T>, never>> : never; type StrictUnion<T> = StrictUnionHelper<T, T> class DynamicProps extends React.Component<StrictUnion<Props>> { render() { return null } } // error now const none_jsx = <DynamicProps type="none" value="This should be an error?" />