如果我想做简单的链接

class MyClass {
   methodA(): this {
      return this
   }
   methodB(): this {
      return this
   }
}

new MyClass()
   .methodA()
   .methodB()

但根据调用函数只返回示例的一部分，如

class MyClass {
   methodA(): this {
      return this
   }
   methodB(): Omit<this, "methodA"> { // <- stop calling `methodA` after `methodB` was called
      // <- what to return?
   }
}

new MyClass()
   .methodA()
   .methodA()
   .methodB()
   .methodB()
   // .methodA() // <- cannot call `methodA` anymore

• 我该怎么做呢？
  当然，我可以

// typescript does not complain about this
   methodB(): Omit<this, "methodA"> {
      return {
         methodB: this.methodB
      } 
   }
   // but typescript would not complain about this either, weird
   methodB(): Omit<this, "methodA"> {
      return this // <- at least it let me do that without throwing 
   }

但是，如果我想排除单个方法/属性，而不是只包含单个方法/属性，该怎么办？

methodB(): Omit<this, "methodA"> {
      const { methodA, ...instance } = this // Could I destructure 'this'? typescript does not complain but eslint throws there would be a parsing error
      return instance
   }