如何在 Typescript 中创建 UUID 模板文字类型？

Question

有没有人使用新的模板文字类型在 Typescript 中成功编写 UUID 类型？

例如： const id:UUID = "f172b0f1-ea0a-4116-a12c-fc339cb451b6"

这家伙试过：UUID Tweet

但类型太复杂：“表达式产生的联合类型太复杂而无法表示。(2590)”：Example

他的类型：

type Alphabetic = 'a' | 'b' | 'c' | 'd' | 'e' | 'f' | 'g' | 'h' | 'i' | 'j' | 'k' | 'l' | 'm' | 'n' | 'o' | 'p' | 'q' | 'r' | 's' | 't' | 'u' | 'v' | 'w' | 'x' | 'y' | 'z'
type Alphanumeric = Alphabetic | Numeric

type Repeat<
    Char extends string,
    Count extends number,
    Joined extends string = ``,
    Acc extends 0[] = []
> = Acc['length'] extends Count ? Joined : Repeat<Char, Count, `${Joined}${Char}`, [0,...Acc]>

type UUIDV4 = `${Repeat<Alphanumeric, 8>}-${Repeat<Alphanumeric, 4>}-${Repeat<Alphanumeric, 4>}-${Repeat<Alphanumeric, 4>}-${Repeat<Alphanumeric, 12>}````

Answer 1

我不确定是否可行，因为 TS 有他自己的限制。考虑这个例子：

type Alphabetic = 'a' | 'b' | 'c' | 'd' | 'e' | 'f' | 'g' | 'h' | 'i' | 'j' | 'k' | 'l' | 'm' | 'n' | 'o' | 'p' | 'q' | 'r' | 's' | 't' | 'u' | 'v' | 'w' | 'x' | 'y' | 'z'
type Numeric = 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0
type StringNumber<T extends number> = `${T}`
type Alphanumeric = Alphabetic | StringNumber<Numeric>

type Result<T extends string> = T extends any ? T : never;

interface Unit {
  value: Alphanumeric
}

type Check = `${Alphanumeric}${Alphanumeric}${Alphanumeric}`

Check 生成超过 46.6K 种可能状态。如果您再添加一个 ${Alphanumeric} 它将达到限制。

我怀疑是否可以在打字稿类型系统中表示。

也许在Tail recursive evaluation of conditional types

之后有可能

我认为可以只验证字符串是否有效，就像我对十六进制表示所做的 here 但它似乎也不起作用。

您仍然可以验证已知的 uuid 文字：

type Alphabetic = 'a' | 'b' | 'c' | 'd' | 'e' | 'f' | 'g' | 'h' | 'i' | 'j' | 'k' | 'l' | 'm' | 'n' | 'o' | 'p' | 'q' | 'r' | 's' | 't' | 'u' | 'v' | 'w' | 'x' | 'y' | 'z'
type Numeric = 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0
type StringNumber<T extends number> = `${T}`
type Alphanumeric = Alphabetic | StringNumber<Numeric>
type Separator = '-'

type UUID = 'f172b0f1-ea0a-4116-a12c-fc339cb451b6'


type Template = [8, 4, 4, 4, 12];

type IsValid<T, Cache extends string[] = []> =
  T extends string ?
  T extends ''
  ? Cache
  : T extends `${infer Char}${infer Rest}`
  ? Char extends Alphanumeric ? IsValid<Rest, [...Cache, Char]>
  : never
  : never
  : never

type Split<S extends string, D extends string> =
  string extends S ? string[] :
  S extends '' ? [] :
  S extends `${infer T}${D}${infer U}` ? [T, ...Split<U, D>] : [S];

type StringLength<T, Cache extends any[] = []> =
  T extends string ?
  T extends ''
  ? Cache['length']
  : T extends `${infer Char}${infer Rest}`
  ? StringLength<Rest, [...Cache, Char]>
  : never
  : never

type Every<T extends any[], Cache extends any[] = []> =
  T extends []
  ? Cache
  : T extends [infer Head, ...infer Rest]
  ? Every<Rest, [...Cache, IsValid<Head>['length']]>
  : never

type IsCorrect<T extends string,> = Every<Split<T, Separator>> extends Template ? true : false

type Test = IsCorrect<'f172b0f1-ea0a-4116-a12c-fc339cb451b6'> // ok
type Test2 = IsCorrect<'f172b0f1-ea0a-4116-a12c'> // false
type Test3 = IsCorrect<'f172b0f1-ea0a-4116-a12c-fc339cb451b6--'> // false
type Test4 = IsCorrect<'f172b0f1-ea0a-4116-a12c-fc339cb451b'> // false
type Test5 = IsCorrect<'f172b0f1-ea0a-4116-a%%c-fc339cb451b'> // false
type Test6 = IsCorrect<'f172b0f1ea0a-4116-a12c-fc339cb451b6'> // false

Answer 2

Yuri Bogolomov 在他的博客上发布了一个解决方案：https://ybogomolov.me/type-level-uuid/。

这是他的解决方案（参见 TypeScript Playground）：

type VersionChar =
    | '1' | '2' | '3' | '4' | '5';

type Char =
    | '0' | '1' | '2' | '3'
    | '4' | '5' | '6' | '7'
    | '8' | '9' | 'a' | 'b'
    | 'c' | 'd' | 'e' | 'f';

type Prev<X extends number> =
    [never, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...never[]][X];

type HasLength<S extends string, Len extends number> = [Len] extends [0]
    ? (S extends '' ? true : never)
    : (S extends `${infer C}${infer Rest}`
        ? (Lowercase<C> extends Char ? HasLength<Rest, Prev<Len>> : never)
        : never);

type Char4<S extends string> = true extends HasLength<S, 4> ? S : never;
type Char8<S extends string> = true extends HasLength<S, 8> ? S : never;
type Char12<S extends string> = true extends HasLength<S, 12> ? S : never;

type VersionGroup<S extends string> = S extends `${infer Version}${infer Rest}`
    ? (Version extends VersionChar
        ? (true extends HasLength<Rest, 3> ? S : never)
        : never)
    : never;

type NilUUID = '00000000-0000-0000-0000-000000000000';

type UUID<S extends string> = S extends NilUUID
    ? S
    : (S extends `${infer S8}-${infer S4_1}-${infer S4_2}-${infer S4_3}-${infer S12}`
        ? (S8 extends Char8<S8>
            ? (S4_1 extends Char4<S4_1>
                ? (S4_2 extends VersionGroup<S4_2>
                    ? (S4_3 extends Char4<S4_3>
                        ? (S12 extends Char12<S12>
                            ? S
                            : never)
                        : never)
                    : never)
                : never)
            : never)
        : never);


const getUser = <S extends string>(id: UUID<S>): void => console.log(id);

getUser('00000000-0000-0000-0000-000000000000'); // ✅  special Nil UUID
getUser('11111111-1111-0111-1111-111111111111'); // ✅  error: version 0 is a special case
getUser('11111111-1111-1111-1111-111111111111'); // ✅  version 1
getUser('11111111-1111-2111-1111-111111111111'); // ✅  version 2
getUser('11111111-1111-3111-1111-111111111111'); // ✅  version 3
getUser('11111111-1111-4111-1111-111111111111'); // ✅  version 4
getUser('11111111-1111-5111-1111-111111111111'); // ✅  version 5
getUser('11111111-1111-6111-1111-111111111111'); // ✅  error: version 6 doesn't exist
getUser('11111111-1111-1111-1111-11111111111');  // ✅  error: invalid format

如何在 Typescript 中创建 UUID 模板文字类型？

How to create a UUID template literal type in Typescript?

uuid

types

typescript