如何键入检查标称ID [英] How to type check nominal-typed IDs

问题描述

我正在尝试此处所述的基于枚举的名义键入方法： https://basarat.gitbooks.io/typescript/docs/tips/nominalTyping.html

I'm trying out the enum-based nominal typing method described here: https://basarat.gitbooks.io/typescript/docs/tips/nominalTyping.html

enum PersonIdBrand {}
export type PersonId = PersonIdBrand & string

interface Person {
  id: PersonId
  firstName: string
  lastName: string
}

我遇到了一个问题，即将类型添加到测试使用的某些工厂方法中。这些帮助程序方法允许创建具有默认值的测试数据，该默认值可以有选择地覆盖：

I ran into an issue adding types to some factory methods used by tests. These helper methods allow creation of test data with defaults that can be selectively overridden:

const makeTestPerson = ({
  id = 'personId' as PersonId,
  firstName = 'Bob',
  lastName = 'Smith'
}: Partial<Person> = {}): Person => ({
  id,
  firstName,
  lastName
})

const person = makeTestPerson({ lastName: 'Ross' })

但是， tsc 会给出错误：

error TS2322: Type 'PersonId' is not assignable to type 'never'.

11   id = 'personId' as PersonId,

如果我改用 id：string 代替，它编译没有任何问题。有什么方法可以使用 PersonId 进行类型检查？

If I instead use id: string instead, it compiles without any problem. Is there any way to make these functions type check using PersonId?

更新，在进行了更多探索之后，我认为此策略存在一个更基本的问题：

Update having explored a bit more, I think there is a more fundamental problem with this strategy:

const maybePersonId: PersonId | undefined = ("personId" as PersonId)

这也将失败，并显示以下内容：

This also fails with:

TS2322: Type 'PersonId' is not assignable to type 'undefined'.

那为什么会失败？当然，应该始终将 X 分配给 X |。未定义？

So why does that fail? Surely an X should always be assignable to X | undefined?

推荐答案

我认为自从编写该代码以来，打字稿已经改变了处理联合的方式，与空的交集（或它被视为空类型的交集）。我找不到能改变行为的PR，但是如果找到它，我会尝试发布它（我找不到PR，但中断发生在2.9）

I think since that code was written typescript has changed the way it handles unions and intersections with empty (or what it perceives as empty type). I am unable to find the PR that changes the behavior but I will try to post it if I find it (I can't find the PR but the break occurs in 2.9)

对于品牌类型，编译器团队使用一个交集，该交集的类型仅包含一个额外的成员，而不是与枚举的交集：

For branded types the compiler team uses an intersection with a type that just contains one extra member not an intersection with an enum:

export type Path = string & { __pathBrand: any };

我只是采用这种方法：

export type PersonId = { __personIdBran: any } & string

interface Person {
  id: PersonId
  firstName: string
  lastName: string
}

const makeTestPerson = ({
  id = 'personId' as PersonId,
  firstName = 'Bob',
  lastName = 'Smith'
}: Partial<Person> = {}): Person => ({
  id,
  firstName,
  lastName
})

const person = makeTestPerson({ lastName: 'Ross' })

这篇关于如何键入检查标称ID的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！