英文:
How to create new interface with Observable's generic types?
问题
我有一个带有一些Observable属性的接口，但接口的类型是未知的。是否有一种方法可以自动创建一个新的接口副本，但不包含Observable
所以这是一个示例：
interface SomeObject
{
value1: Observable<string>;
value2: boolean;
value3: Observable<number>;
}
是否有一种方法可以从中创建一个新的接口，结果如下所示？
interface SomeObjectWithoutObservable
{
value1: string;
value2: boolean;
value3: number;
}
也许如果有一种类似于这样的方法？
interface SomeObjectWithoutObservable extends RemoveObservable<SomeObject>
{}
只是不确定RemoveObservable会是什么样子。
英文:
Say I have an interface with some Observable properties, but the type of interface is unknown. Is there a way I can automatically create a new copy of the interface but without Observable<T>?
So here is an example:
interface SomeObject
{
value1: Observable<string>;
value2: boolean;
value3: Observable<number>;
}
Is there a way I can create a new interface from that that would result in this?
interface SomeObjectWithoutObservable
{
value1: string;
value2: boolean;
value3: number;
}
Maybe if there was a way to do it similar to this?
interface SomeObjectWithoutObservable extends RemoveObservable<SomeObject>
{}
Just not sure what RemoveObservable would look like.
答案1
得分: 2
我相信你可以通过这种方法实现它：
type RmObservable<T extends object> = {
[key in keyof T]: T[key] extends Observable<infer R> ? R : T[key];
};
interface SomeObjectWithoutObservable extends RmObservable<SomeObject> { }
演示：https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgPICMDO0Bud0A2EAPACoB8yA3gFDLJ4ECuEAXMqTQL402iSxEKAMoB7ALYQMAKwgIw1OgzjMIARnYZsUPIRKYwUUAHNyAbiWMWAJnbpRoonBAX6ViAGZNWXPiLEQJnF0aHNuXjAATwAHFAAlcS1fPTJkCAAPSBAAE0xkUXRZeUoAXkV6AG0AawhI5FBkGsjRGA4AXXZSatq2tMyIHLyknT8SUBhoZDjKAH4p5E7uyLaLLgs+cGh4JGQxSRk5MAB1YDAAC1EmMGHdIj6s3KnEnxGUvalCw8oqZB4gA
英文:
I believe you can achieve it by this method:
type RmObservable<T extends object> = {
[key in keyof T]: T[key] extends Observable<infer R> ? R : T[key];
};
interface SomeObjectWithoutObservable extends RmObservable<SomeObject> { }
答案2
得分: 0
感谢Tachibana！这有效。我只想添加另一个答案以扩展您的答案。我忘了提到，我总是在Observables属性名称的末尾添加$。像这样：
interface SomeObject
{
value1: Observable<string>;
value2: boolean;
value3: Observable<number>;
}
所以对于这个，我也想要去掉结果中的$。所以我想出了这个：
type Replace<
T extends string,
S extends string,
D extends string,
A extends string = ''
> = T extends `${infer L}${S}${infer R}` ? Replace<R, S, D, `${A}${L}${D}`> : `${A}${T}`;
type RmObservable<T extends object> = {
[key in keyof T as `${T[key] extends Observable<any> ? Replace<key, '$', ''> : key}`]: T[key] extends Observable<infer R> ? R : T[key];
};
interface SomeObjectWithoutObservable extends RmObservable<SomeObject> { }
感谢这个答案提供了Replace功能：https://stackoverflow.com/a/71353081/8916441
免责声明：我知道这不是完美的，因为这将替换属性名称中的任何部分的$，而不仅仅是在末尾。但我认为在我的项目中不需要担心这一点。
英文:
Thanks Tachibana! That works. I just wanted to add another answer to expand on yours. I forgot to mention that I always have a $ at the end of my property names for Observables. Like this:
interface SomeObject
{
value1$: Observable<string>;
value2: boolean;
value3$: Observable<number>;
}
So for this I also wanted to remove the $ in the result. So I came up with this:
type Replace<
T extends string,
S extends string,
D extends string,
A extends string = ''
> = T extends `${infer L}${S}${infer R}` ? Replace<R, S, D, `${A}${L}${D}`> : `${A}${T}`;
type RmObservable<T extends object> = {
[key in keyof T & string as `${T[key] extends Observable<any> ? Replace<key, '$', ''> : key}`]: T[key] extends Observable<infer R> ? R : T[key];
};
interface SomeObjectWithoutObservable extends RmObservable<SomeObject> { }
Credit to this answer for the Replace functionality: https://stackoverflow.com/a/71353081/8916441
Disclaimer: I know this is not perfect, since this would replace $ in ANY part of the property name, not just at the end. But I don't think I need to worry about that in my project.
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