Is it possible to remove an inherited field/method in a child class/interface?

问题

Something along these lines:

interface A {
  a: number;
  x: any;
}

interface B extends A {
  b: number;
}

interface C {
  a: number;
  b: number;
}

So the B would be equal to C (omitting field x but still extending A). Is it possible? If so, how?

回答1:

It is impossible to remove an inherited field/method of an interface in TypeScript.

But you can overcome this via interfaces reengineering:

1. Extract 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 will be castable to BaseA.

2. Use optional field

   interface A {
     a: number;
     x?: any;
   }
   
   interface B extends A {
     b: number;
   }
   
   interface C extends A {
     b: number;
   }
   

I'm sure there are other ways depending on the certain task context.

回答2:

here is another hacky way to deal with deleting child interface members:

interface Base {
  a: number;
  x?: any;
}

interface Child extends Base {
  x?: undefined;
}

this way, either the field x should not exist at all or if it does, it must be set to undefined which is as good as non-existent in most cases (unless you are trying to enumerate class members or do things like that).

来源：https://stackoverflow.com/questions/41671281/is-it-possible-to-remove-an-inherited-field-method-in-a-child-class-interface