英文:
Create function in interface with name from const in Typescript
问题
export const Events = {user_updated: "user:updated",empty: "empty"}interface ServerToClientEvents {[Events.empty]: () => void;[Events.user_updated]: (user: User) => void;}
英文:
Is it possible to create function in interface with name from const? I have const where I have all event names and I want to create ServerToClientEvents interface for SocketIO.
I can write something like this and it will work
interface A {"funcName": () => void}
But can I create functions in this interface with names from FunNames?
export const Events = {user_updated: "user:updated",empty: "empty"}interface ServerToClientEvents {Events.empty: () => void;Events.user_updated: (user: User) => void;}
答案1
得分: 2
如果Events不是动态的，也许const断言可以胜任任务。
export const Events = {user_updated: 'user:updated',empty: 'empty',} as const; //const断言interface User {id: string;}interface ServerToClientEvents {[Events.empty]: () => void;[Events.user_updated]: (user: User) => void;}
英文:
If Events isn't dynamic, maybe const assertion can do the job.
export const Events = {user_updated: 'user:updated',empty: 'empty',} as const; //const assertioninterface User {id: string;}interface ServerToClientEvents {[Events.empty]: () => void;[Events.user_updated]: (user: User) => void;}
答案2
得分: 1
你可以使用一个mapped type和一个嵌套的conditional type，测试我们映射的键是否与template literal type匹配（这样我们可以检测是否需要user参数）：
export const Events = {user_updated: "user:updated",empty: "empty",};type ServerToClientEvents = {[Key in keyof typeof Events]: Key extends `user_${string}` ? (user: User) => void : () => void;};const goodExample: ServerToClientEvents = {user_updated(user: User) {},empty() {},};const badExample: ServerToClientEvents = {//^^^^^^^^^^^−−−−− Property 'empty' is missing in type '{ user_updated(user: User): void; }' but required in type 'ServerToClientEvents'.ts(2741)user_updated(user: User) {},};
在那里，我假设user_前缀应该表示函数接受一个User对象，根据问题中显示的代码，但你当然可以根据需要进行调整。
使用给定的Events常量，该映射类型定义了这个结构：
{user_updated: (user: User) => void;empty: () => void;}
为了提高可读性，你可以拆分条件类型：
type ServerFunctionType<Key extends string> =Key extends `user_${string}`? (user: User) => void: () => void;type ServerToClientEvents = {[Key in keyof typeof Events]: ServerFunctionType<Key>;};
英文:
You can do it with a mapped type and a nested conditional type testing the key we're mapping to see if it matches a template literal type (so we detect whether we need the user parameter):
export const Events = {user_updated: "user:updated",empty: "empty",};type ServerToClientEvents = {[Key in keyof typeof Events]: Key extends `user_${string}` ? (user: User) => void : () => void;};const goodExample: ServerToClientEvents = {user_updated(user: User) {},empty() {},};const badExample: ServerToClientEvents = {//^^^^^^^^^^^−−−−− Property 'empty' is missing in type '{ user_updated(user: User): void; }' but required in type 'ServerToClientEvents'.ts(2741)user_updated(user: User) {},};
I've assumed there that the user_ prefix should mean the function accepts a User object, given the code shown in the question, but of course you can tweak that as necessary.
With the given Events constant, that mapped typ defines this structure:
{user_updated: (user: User) => void;empty: () => void;}
For readability, you might split out the conditional type:
type ServerFunctionType<Key extends string> =Key extends `user_${string}`? (user: User) => void: () => void;type ServerToClientEvents = {[Key in keyof typeof Events]: ServerFunctionType<Key>;};
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