[英]TypeScript: generically infer union type member based on a string literal property
TypeScript (v3.2.2) allows me to define a union of interfaces, each with a unique string literal property which can be used as a type guard, eg 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.`);
}
}
I want to write a function isDeviceType
in a generic way:我想以通用方式编写 function
isDeviceType
:
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
However, the way I have defined the DeviceOf
type is pretty verbose since it lists every single type within the union: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一个类型:
type DeviceOf<Type extends Device['type']> =
Type extends Laptop['type'] ? Laptop :
Type extends Desktop['type'] ? Desktop :
Type extends Phone['type'] ? Phone :
never;
Is there a more concise way to define DeviceOf
?有没有更简洁的方法来定义
DeviceOf
? I have tried these: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'.
My impression is that error TS1338 is the limiting factor, and so it's impossible to define DeviceOf
in a generic way in the current version of TypeScript.我的印象是错误 TS1338 是限制因素,因此不可能在当前版本的 TypeScript 中以通用方式定义
DeviceOf
。
Got it.知道了。 You have to apply "if" twice, once for create
infer
type and second to check if infer
type extends device.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。 Only in branch D extends Device
you will be able u use D['type']
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'>;
Found an alternative way, using just conditional types without the infer
keyword:找到了另一种方法,只使用没有
infer
关键字的条件类型:
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基于 Ryan Cavanaugh 的评论: https : //github.com/Microsoft/TypeScript/issues/17915#issuecomment-413347828
Typescript 2.8 supports the Extract<Type, Union>
utility, which makes this even simpler to implement. Typescript 2.8 支持
Extract<Type, Union>
实用程序,这使得实现起来更加简单。
type Result = Extract<Device, { type: 'Laptop' }>;
From the Typescript documentation here :来自此处的 Typescript 文档:
Constructs a type by extracting from
Type
all union members that are assignable toUnion
.通过从
Type
中提取可分配给Union
的所有联合成员来构造一个类型。
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