Question

当我尝试将钩子传递给功能组件时，出现此错误：

override fun onActivityResult(requestCode: Int, resultCode: Int, data: Intent?) {
    if(requestCode == 123 && resultCode == Activity.RESULT_OK){
         tvCode.setText(data.getExtras().getString("inputText",""));
    } else {
        super.onActivityResult(requestCode, resultCode, data)
    }
}

Type '() => void' is not assignable to type 'FunctionComponent<InputProps>'.
  Type 'void' is not assignable to type 'ReactElement<any, string | ((props: any) => ReactElement<any, string | ... | (new (props: any) => Component<any, any, any>)> | null) | (new (props: any) => Component<any, any, any>)> | null'.

Answer 1

您的功能组件签名（React.FunctionComponent）与您的功能componenet实现（（）=> {}）不匹配

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如果查看FunctionComponent接口，您会注意到它必须实现像（props：P＆{children ?: ReactNode}，context ?: any）这样的函数：ReactElement | null;

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与Typescript反应：无法将函数prop传递给子组件

1 个答案: