举个例子
interface Person<TName> {
name: TName;
}
type People =
| Person<"Levi">
| Person<"Julien">
type FilteredPersonBase<TPerson extends People> = TPerson["name"] extends "Levi" ? TPerson : never;
type FilteredPerson = FilteredPersonBase<People>; // never
type FilteredPersonBase2<TPerson extends People> = TPerson extends { name: "Levi" } ? TPerson : never;
type FilteredPerson2 = FilteredPersonBase2<People>; // Person<"Levi">
我希望这两个 FilteredPerson和 FilteredPerson2解决Person<"Levi"> ，但他们没有。为什么在类型约束中使用索引访问运算符的解析方式与使用内联类型不同？
最佳答案
我同意第二个应该解决人 Person<"Levi"> ，但让我们看看这是如何发生的:
FilteredPersonBase2<People>
// FilteredPersonBase2 is a distributive conditional type since
// TPerson appears nakedly in the condition of the conditional type
=> FilteredPersonBase2<Person<"Levi">> | FilteredPersonBase2<Person<"Julien">>
// The conditional type is applied to each member of the union
=> (Person<"Levi"> extends { name: "Levi" } ? Person<"Levi"> : never) | (Person<"Julien"> extends { name: "Levi" } ? Person<"Julien"> : never)
// The first application return Person<"Levi">, the second resolves to never
=> Person<"Levi"> | never
// Never is removed in a union
=> Person<"Levi">
这样做的原因是 distributive behavior of conditional types
但是条件类型只分布在裸类型参数上。在第一个示例中，条件结束 TPerson["name"]这不是裸类型参数，因此不会发生分配。展开我们得到的条件类型:
FilteredPersonBase<People> =>
// No distribution
People["name"] extends "Levi" ? People : never =>
// People["name"] just resolves to the union
"Levi" | "Julien" extends "Levi" ? People : never =>
// The union does not extend a constituent of the union
never
由于没有发生分发，TPerson["name"]只是解析为 "Levi" | "Julien"联合不会扩展成分。反之亦然 ("Levi" extends TPerson["name"])，但这只会解析为联合 (Person)，因为联合的成分是联合的子类型。
只是为了好玩，您可以强制分发 TPerson使用始终为真的条件，如 TPerson extends TPerson :
type FilteredPersonBase<TPerson extends People> = TPerson extends TPerson ? TPerson["name"] extends "Levi" ? TPerson : never : never;
type FilteredPerson = FilteredPersonBase<People>; // Person<"Levi">
关于TypeScript 索引访问类型约束行为异常，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/57040532/