Typescript TS 条件类型无法识别条件参数

简而言之，我需要一个返回either类型"Person"或类型"PersonWithSurname"的函数.这与条件类型一起工作得很好.然而，我正在努力在没有错误的情况下构建结果.

interface Person {
  name: string;
}

type PersonWithSurname = Person & { 
  surname : string 
};

type PersonNameConditional<T> = T extends true 
  ? PersonWithSurname 
  : Person; 

function getPerson<T extends boolean>(returnSurname: withSurname
  ): PersonNameConditional<T> {
    if (returnSurname) {
      return { name: "John", surname: "Wic-k" } as PersonNameConditional<withSurname>
    } else {
      return { name: "John" } as PersonNameConditional<withSurname>
    }
}

// All good here ✅

上面的方法运行得很好，甚至IntelliSense也会根据收到的参数向我显示返回的类型，如下所示.

const result = getPerson(true) // { name: "John", surname: "Wic-k" }: PersonWithSurname
const result = getPerson(false) // { name: "John" }: Person

// All good here ✅

然而，在这里我遇到了一个问题，TS无法从响应中识别surname.

const requestedSurname: boolean = req.query.withSurname

let result = getPerson(requestedSurname)

if ("surname" in result)
  // All good here  ✅
  result.surname = result.surname.replace(/-/g, '')

if (result.surname)
  // ❌ Property "surname" does not exist on type { name : string } | PersonWithSurname
  result.surname = result.surname.replace(/-/g, '')

// the ideal usage
if (requestedSurname) 
  // ❌ Property "surname" does not exist on type { name : string } | PersonWithSurname
  result.surname = result.surname.replace(/-/g, '')

return result;

难道不能使用条件变量吗？