看起来“OneOf”的意思是“必须匹配”exactlyone”,而“AnyOf”表示“必须匹配at least原来“至少一个”是一个比较基本的概念,对应的是union https://en.wikipedia.org/wiki/Union_(set_theory)手术 (”包含或 https://en.wikipedia.org/wiki/Logical_disjunction”)代表的是|
象征。因此,您所提出的问题的答案只是:
type AnyOfSample = HasName | HasEmail // Union type
与交集的进一步并集不会改变接受的值:
type AnyOfSample = HasName | HasEmail | (HasName & HasEmail)
因为联合只能添加元素,并且所有元素HasName & HasEmail
已经存在于HasName | HasEmail
。观察:
type HasName = { name: string }
type HasEmail = { email: string };
type AnyOfIntersection = HasName | HasEmail | (HasName & HasEmail)
type AnyOfWithout = HasName | HasEmail
const aI: AnyOfIntersection = { name: "", email: "" }; // of course
const aW: AnyOfWithout = { name: "", email: "" }; // also accepted
如果您要使用,您可能需要保留交叉点the in缩小值的运算符 https://www.typescriptlang.org/docs/handbook/2/narrowing.html#the-in-operator-narrowing,这做出了技术上不正确但通常有用的假设,即如果密钥是未知存在于一个类型中,那么它是not展示:
function processAnyOf(aI: AnyOfIntersection, aW: AnyOfWithout) {
if (("name" in aI) && ("email" in aI)) {
aI; // (HasName & HasEmail)
}
if (("name" in aW) && ("email" in aW)) {
aW; // never
}
}
当然,这意味着您的定义不正确OneOfSample
。这个操作更像是一个析取联合 https://en.wikipedia.org/wiki/Symmetric_difference (“异或” https://en.wikipedia.org/wiki/Exclusive_or),尽管不完全是因为当您有三个或更多集合时,析取并的通常定义意味着“匹配奇数”,这不是您想要的。顺便说一句,我找不到我们在这里讨论的析取并类型的广泛使用的名称,尽管这里有一个有趣的论文 http://www.sfu.ca/%7Ejeffpell/papers/IGPLTernaryExclOr.pdf那个讨论它。
那么,我们如何在 TypeScript 中表示“完全匹配”呢?这并不简单,因为它最容易构建为negation https://github.com/Microsoft/TypeScript/issues/4196 or 减法 https://github.com/Microsoft/TypeScript/issues/4183类型,TypeScript 目前无法做到这一点。也就是说,你想说的是:
type OneOfSample = (HasName | HasEmail) & Not<HasName & HasEmail>; // Not doesn't exist
但没有Not
这在这里有效。因此,你所能做的就是某种解决方法……那么有什么可能呢?你can告诉 TypeScript 类型可能没有特定属性。例如类型NoFoo
可能没有foo
key:
type ProhibitKeys<K extends keyof any> = {[P in K]?: never};
type NoFoo = ProhibitKeys<'foo'>; // becomes {foo?: never};
And you can使用条件类型获取键名称列表并从另一个列表中删除键名称(即减去字符串文字):
type Subtract = Exclude<'a'|'b'|'c', 'c'|'d'>; // becomes 'a'|'b'
这可以让您执行如下操作:
type AllKeysOf<T> = T extends any ? keyof T : never; // get all keys of a union
type ProhibitKeys<K extends keyof any> = {[P in K]?: never }; // from above
type ExactlyOneOf<T extends any[]> = {
[K in keyof T]: T[K] & ProhibitKeys<Exclude<AllKeysOf<T[number]>, keyof T[K]>>;
}[number];
在这种情况下,ExactlyOneOf
需要一个类型的元组,并将表示该元组中每个元素的并集,明确禁止来自其他类型的键。让我们看看它的实际效果:
type HasName = { name: string };
type HasEmail = { email: string };
type OneOfSample = ExactlyOneOf<[HasName, HasEmail]>;
如果我们检查OneOfSample
有了 IntelliSense,它是:
type OneOfSample = (HasEmail & ProhibitKeys<"name">) | (HasName & ProhibitKeys<"email">);
这就是说“要么HasEmail
没有name
财产,或HasName
没有email
财产。有效吗?
const okayName: OneOfSample = { name: "Rando" }; // okay
const okayEmail: OneOfSample = { email: "[email protected] /cdn-cgi/l/email-protection" }; // okay
const notOkay: OneOfSample = { name: "Rando", email: "[email protected] /cdn-cgi/l/email-protection" }; // error
看起来是这样。
元组语法允许您添加三种或更多类型:
type HasCoolSunglasses = { shades: true };
type AnotherOneOfSample = ExactlyOneOf<[HasName, HasEmail, HasCoolSunglasses]>;
这检查为
type AnotherOneOfSample = (HasEmail & ProhibitKeys<"name" | "shades">) |
(HasName & ProhibitKeys<"email" | "shades">) |
(HasCoolSunglasses & ProhibitKeys<"email" | "name">)
正如您所看到的,它正确地分配了禁止的密钥。
还有其他方法可以做到这一点,但这就是我要继续的方式。这是一种解决方法,而不是一个完美的解决方案,因为有些情况它无法正确处理,例如具有相同键的两种类型,其属性是不同类型:
declare class Animal { legs: number };
declare class Dog extends Animal { bark(): void };
declare class Cat extends Animal { meow(): void };
type HasPetCat = { pet: Cat };
type HasPetDog = { pet: Dog };
type HasOneOfPetCatOrDog = ExactlyOneOf<[HasPetCat, HasPetDog]>;
declare const abomination: Cat & Dog;
const oops: HasOneOfPetCatOrDog = { pet: abomination }; // not an error
在上文中,ExactlyOneOf<>
无法递归到的属性pet
财产,以确保它不是两个Cat
and a Dog
。这个问题可以解决,但它开始变得比您想要的更复杂。还有其他边缘情况。这取决于你需要什么。
Playground 代码链接 https://www.typescriptlang.org/play?#code/C4TwDgpgBAEghgZwHJwLbQLxQN5QHZoQBcUCwATgJZ4DmUAvgFCiSyICiqclANlFrghdeJMlVoMA3I2bhoAQTwgA8gDMAymjA9MbZISgAfPZ258A9OaNQAFPH3ooAMhPCeAShksFStQEk8YAhyBAgAY2BKAHs8fj0UR2N7U15rO0QE6Bdkt09vKEUVVQB1SmAACyiAV2A4+0zrHLMZMJiyKDg-EkL-QODQiOjYgXxCEgAicYAaKCEzCfGpKEsoKNUoVqqQiEZWvHa4Yu7fErLKmrjcAnQFmbmRKEmllbgeBCiOsLCIMCCAExkqiqeEGMSgYHIUW+CAQPVUNk6xyKASC21BeBmhyRalKFWqwHcOEYUBJUEo6xsNnG1wgi2oHT8hKcLip9x4dNinXchOwxNJ-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-gBeEEdJxFBXgVSoMe5DuTMflQEFw6huGjD0D2yWpcEQ4PmZdoQA6Wg6J+KaWdZlXET2gboDMTS8O50jZqwTUaDVLodXwLhheuhF+tShDjKGhLGK6QZ5gW4UbtJE0ajN5V7O0UTNnAICZCQfWOs63Y0hMABKykxos96Hjte27G0tT3SAKQ8IdtnHbVIxshM5BXVEAACqYDQAdK0qA3chL2PfstR4L6yi7SAH1Vf1TrfTgow3I8l0NlE0yzG4f0A8D9UYxA4OnlDh7BJC5CZvk9gAMKLjwz60DwiChE2IwIOUcD-ggJAUFU0D3mZii+uUwRHQNdUNTZaN+i1GSEO1Ia8BrCCs1E7PAjQXMwnGXWZuYABU+RSxUsuffLWCim9oVLWNq2TegizGOM-OC3G02zeatiteKo0RVW6puJ7jw+0L-tGIH-Kirr+uc9zcbO6uy3jRHZhR+7tIzeb5iZv+P5wOQ0BlzCBR4JQXB8LgOg0ML+ApWl06l1zFcbEbTYACJRHQcmOTRih168WNFuXzY2O4JDGlElB-B34Rd5XvdQMzcDJnWI9NmP9dY+gUQAO6z-Pi-LxLTOILmXFb7UIyQMAJAP9ON8IHfwAD3QT9cSQP9345nsH1L+D9lDkEAVgCyERGq2RVvYMB29tZfx-qbTu5dK5PSMqeag28hiv23l6H+0gtrPSiGAFuIDbJIOABAqBWNn4kDgMZVAeDIhgluoaX0HZZjkHpkAA