TypeScript：根据字符串文字属性一般推断联合类型成员

Question

TypeScript (v3.2.2) 允许我定义接口的联合，每个接口都有一个唯一的字符串文字属性，可以用作类型保护，例如

type Device = Laptop | Desktop | Phone;

interface Laptop {
  type: 'Laptop';
  countDriveBays: number;
  hasTouchScreen: boolean;
}

interface Desktop {
  type: 'Desktop';
  countDriveBays: number;
}

interface Phone {
  type: 'Phone';
  hasTouchScreen: boolean;
}

function printInfo(device: Device) {
  if (device.type === 'Laptop') {
    // device: Laptop
    console.log(
      `A laptop with ${device.countDriveBays} drive bays and ${
        device.hasTouchScreen ? 'a' : 'no'
      } touchscreen.`,
    );
  } else if (device.type === 'Desktop') {
    // device: Desktop
    console.log(`A desktop with ${device.countDriveBays} drive bays.`);
  } else {
    // device: Phone
    console.log(`A phone with ${device.hasTouchScreen ? 'a' : 'no'} touchscreen.`);
  }
}

我想以isDeviceType通用方式编写一个函数：

const isDeviceType = <T extends Device['type']>(type: T) => {
  return (device: Device): device is DeviceOf<T> => device.type === type;
}

// e.g.
const isPhone = isDeviceType('Phone');
isPhone({ type: 'Phone', hasTouchScreen: true }); // true

但是，我定义DeviceOf类型的方式非常冗长，因为它列出了联合中的每个类型：

type DeviceOf<Type extends Device['type']> =
  Type extends Laptop['type'] ? Laptop :
  Type extends Desktop['type'] ? Desktop :
  Type extends Phone['type'] ? Phone :
  never;

有没有更简洁的定义方式DeviceOf？我试过这些：

type DeviceOf<Type extends Device['type']> =
  (infer D)['type'] extends Type ? D : never;

// TS2536: Type '"type"' cannot be used to index type 'D'.
// TS1338: 'infer' declarations are only permitted in the 'extends' clause of a conditional type.
// TS6133: 'D' is declared but its value is never read.

type DeviceOf<Type extends Device['type']> =
  (infer D) extends Device
    ? D['type'] extends Type
    ? D
    : never
    : never;

// TS1338: 'infer' declarations are only permitted in the 'extends' clause of a conditional type.
// TS6133: 'D' is declared but its value is never read.
// TS2304: Cannot find name 'D'.

我的印象是错误 TS1338 是限制因素，因此DeviceOf在当前版本的 TypeScript 中不可能以通用方式定义。

Answer 1

Found an alternative way, using just conditional types without the infer keyword:

type FindByType<Union, Type> = Union extends { type: Type } ? Union : never;
type DeviceOf<Type extends Device['type']> = FindByType<Device, Type>;

type Result = DeviceOf<'Laptop'>;

Based on Ryan Cavanaugh's comment here: https://github.com/Microsoft/TypeScript/issues/17915#issuecomment-413347828

Answer 2

知道了。您必须应用“if”两次，一次用于创建infer类型，第二次用于检查infer类型是否扩展设备。只有在分支D extends Device你才能使用D['type']

type DeviceOf<Type extends Device['type']> =
  Device extends (infer D) ?
  D extends Device ?
  D['type'] extends Type ? D : never : never : never;

type Result = DeviceOf<'Laptop'>;

操场