处理联合类型时出现打字稿错误

Question

我有以下函数，将三种类型之一的参数作为参数。通过类型比较，我选择了正确的响应形状，如下所示。

const buildAddress = (
  address: VerifiedStandardAddress | VerifiedExtendedAddress | Address
): Record<string, unknown> => {
  if (typeof address !== Address.name) {
    return {
      city: address.city,
      line1: address.address_line_1,
      line2: address.address_line_2,
      last_line: address.address_last_line,
      postal_code: address.postal,
      postal_ext: address.postal_ext,
      state: address.region
    }
  }

  return {
    city: address.city,
    line1: address.line1,
    line2: address.line2,
    postal_code: address.postal_code,
    state: address.state
  }
}

但是，打字稿编译器会抱怨，例如，

Property 'state' does not exist on type 'VerifiedExtendedAddress'

。它是正确的。此属性存在于类型

Address

上。

有没有办法避免这种情况，而不是将这段代码分成两个函数？

Answer 1

要么

typeof address !== Address.name

并不总是匹配

VerifiedExtendedAddress

，要么 Typescript 的智能无法推断隐式类型，因为关系太复杂。

在第一种情况下，您需要编辑 if 子句，使其始终与 VerifiedExtendedAddress 匹配。 在第二种情况下，您需要更强的类型保护。一个非常明确的类型保护函数可能如下所示：

function isFish(pet: Fish | Bird): pet is Fish {
    return (pet as Fish).swim !== undefined;
}

然后您可以调用 isFish(...) 并确保捕获“鱼案例”。