typescript - 基于可选通用的接口(interface)中的强制键

Question

Typescript 中有没有办法让接口(interface)在传递泛型时具有强制键？

我正在寻找一种方法，仅当将泛型传递给接口(interface)时才能够为接口(interface)中的键定义类型。

例如。

interface IExample {
  foo: string
}
​
/* You can't declare 2 interfaces of the same name, but this shows the structure I am aiming for */
interface IExample<T> {
  foo: string,
  bar: T
}
​
/* Allowed */
const withoutBar: IExample {
  foo: 'some string'
}
​
/* Not allowed, as I've passed in a generic for Bar */
const withoutBar: IExample<number> {
  foo: 'some string'
}
​
/* Allowed */
const withBar: IExample<number> {
  foo: 'some string',
  bar: 1
};
​
/* Not allowed as I have not passed in a generic for Bar */
const withBar: IExample {
  foo: 'some string',
  bar: 1 // Should error on "bar" as I have not passed in a generic
};

Answer 1

您可以使用条件类型的类型别名。

type IExample<T = void> = T extends void ?  {
  foo: string
} : {
  foo: string,
  bar: T
}
​​
/* Allowed */
const withoutBar: IExample = {
  foo: 'some string'
}
​
/* Not allowed, as I've passed in a generic for Bar */
const withoutBar: IExample<number> = {
  foo: 'some string'
}
​
/* Allowed */
const withBar: IExample<number> = {
  foo: 'some string',
  bar: 1
};
​
/* Not allowed as I have not passed in a generic for Bar */
const withBar: IExample = {
  foo: 'some string',
  bar: 1 // Should error on "bar" as I have not passed in a generic
};

Playground