如何声明AuthType类型，使navigateTo排除accepted指定的枚举键？

你需要利用TypeScript的Exclude工具类型，从AuthStatus枚举中移除被指定为accepted的泛型参数T，以此约束navigateTo的键只能是剩下的枚举成员。

修正后的类型声明 #

enum AuthStatus {
  UNAUTHENTICATED,
  ONBOARDING,
  AUTHENTICATED
}

type AuthType<T extends AuthStatus> = {
  accepted: T,
  // 用Exclude排除掉accepted对应的枚举值，只保留剩余成员作为键
  navigateTo: Record<Exclude<AuthStatus, T>, string>
}

验证示例 #

给每个实例指定明确的泛型参数，TypeScript会自动校验navigateTo的键是否符合要求：

// accepted为UNAUTHENTICATED，navigateTo自动排除该值
const exampleA: AuthType<AuthStatus.UNAUTHENTICATED> = {
  accepted: AuthStatus.UNAUTHENTICATED,
  navigateTo: {
    [AuthStatus.ONBOARDING]: '/onboarding',
    [AuthStatus.AUTHENTICATED]: '/',
  }
}

// accepted为ONBOARDING，navigateTo自动排除该值
const exampleB: AuthType<AuthStatus.ONBOARDING> = {
  accepted: AuthStatus.ONBOARDING,
  navigateTo: {
    [AuthStatus.UNAUTHENTICATED]: '/login',
    [AuthStatus.AUTHENTICATED]: '/',
  }
}

// accepted为AUTHENTICATED，navigateTo自动排除该值
const exampleC: AuthType<AuthStatus.AUTHENTICATED> = {
  accepted: AuthStatus.AUTHENTICATED,
  navigateTo: {
    [AuthStatus.UNAUTHENTICATED]: '/login',
    [AuthStatus.ONBOARDING]: '/onboarding',
  }
}

错误校验示例 #

如果navigateTo中包含了accepted对应的枚举值，TypeScript会直接抛出类型错误：

// 此处会触发类型错误：navigateTo中包含了accepted的UNAUTHENTICATED
const badExample: AuthType<AuthStatus.UNAUTHENTICATED> = {
  accepted: AuthStatus.UNAUTHENTICATED,
  navigateTo: {
    [AuthStatus.UNAUTHENTICATED]: '/login',
    [AuthStatus.ONBOARDING]: '/onboarding',
  }
}

内容的提问来源于stack exchange，提问作者W.S.