扩展 Typescript 接口时推断泛型参数

Question

I have an interface which requires a generic variable:我有一个需要通用变量的interface ：

interface Action<T extends string> {
   type: T
}

I am able to extend it when I set the value of the generic:当我设置泛型的值时，我可以扩展它：

interface SomeAction extends Action<"something"> {
}

I would like to infer the generic automatically, but I get an error when I try this:我想自动推断泛型，但是当我尝试这样做时出现错误：

interface SomeAction extends Action {
   type: "something"
}

Error TS2314: Generic type 'Action' requires 1 type argument(s).错误 TS2314：通用类型“操作”需要 1 个类型参数。

I'd like the second example to infer its type automatically.我希望第二个示例自动推断其类型。 Is this possible?这可能吗？

Answer 1

Just specify default value for T and it should work:只需为T指定默认值，它应该可以工作：

interface Action<T extends string = string> {}