Hello TypeScript experts.
I have the following code but I have to repeat the interface properties in the class otherwise I get:
Class incorrec
There is no built-in support for this.
We can however use a function that returns a class as the base type of our class. This function can lie a little bit and claim it implements the interface. We could also pass in some defaults for the members if necessary.
interface INavigation {
Id: number;
AppId: number;
NavId: number;
Name: string;
ParentId: string;
PageURL: string;
Position: string;
Active: string;
Desktop: string;
Tablet: string;
Phone: string;
RoleId: string;
Target: string;
}
function autoImplement<T>(defaults?: Partial<T>) {
return class {
constructor() {
Object.assign(this, defaults || {});
}
} as new () => T
}
class Navigation extends autoImplement<INavigation>() {
constructor(navigation: any) {
super();
// other init here
}
}
If we want to have a base class, things get a bit more complicated since we have to perform a bit of surgery on the base type:
function autoImplementWithBase<TBase extends new (...args: any[]) => any>(base: TBase) {
return function <T>(defaults?: Partial<T>): Pick<TBase, keyof TBase> & {
new(...a: (TBase extends new (...o: infer A) => unknown ? A : [])): InstanceType<TBase> & T
} {
return class extends base {
constructor(...a: any[]) {
super(...a);
Object.assign(this, defaults || {});
}
} as any
}
}
class BaseClass {
m() { }
foo: string
static staticM() { }
static staticFoo: string
}
class Navigation extends autoImplementWithBase(BaseClass)<INavigation>() {
constructor(navigation: any) {
super();
// Other init here
}
}
Navigation.staticFoo
Navigation.staticM
new Navigation(null).m();
new Navigation(null).foo;
This is now possible in Typescript using class/interface merging.
interface Foo {
a: number;
}
interface Baz extends Foo { }
class Baz {
constructor() {
console.log(this.a); // no error here
}
}
https://github.com/Microsoft/TypeScript/issues/340#issuecomment-184964440
Class declarations should explicitly implement interfaces.