Here's a scenario:
interface A {
a: number;
x: any;
}
interface B extends A {
b: number;
}
interface C {
a: number;
b: number;
}
Can we make B equal to C (excluding the x field but still extending A)? If it is possible, then how can we achieve it?
In TypeScript, removing an inherited field or method from an interface is not possible.
However, you can overcome this limitation by reengineering interfaces:
Extract a base interface
interface BaseA {
a: number;
}
interface A extends Base A {
x: any;
}
interface B extends A {
b: number;
}
interface C extends BaseA {
b: number;
}
Both C and B can be cast to BaseA.
Utilize optional fields
interface A {
a: number;
x?: any;
}
interface B extends A {
b: number;
}
interface C extends A {
b: number;
}
There may be other approaches depending on the specific context of the task.
If you're looking for a workaround to handle the deletion of child interface members, here's a clever solution:
interface Base {
a: number;
x?: any;
}
interface Child extends Base {
x?: undefined;
}
By using this approach, the x field either shouldn't exist at all, or if it does, it must be explicitly set to undefined. This effectively eliminates the field in most cases, unless you specifically need to access class members or perform similar operations.
To achieve this, you can utilize typescripts omit type-keys feature. It's important to note that in typescript, you have the ability to extend multiple interfaces simultaneously.
Typescript omit:
Creates a type by selecting all properties from Type and then removing Keys (string literal or union of string literals).
Using the example provided:
interface A {
a: number;
x: any;
}
interface B extends A {
b: number;
}
interface C extends Omit<A, 'x'>, Omit<B, 'x'> {}
The result:
interface C {
a: number;
b: number;
}
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