Console log an interface without some object

Question

Do you know how can I console log the interface example without the m and n? Below the two interface is in one file.

export interface example {
  readonly a: string;
  readonly b: F | undefined;
  readonly c: G | undefined;
  readonly d?: boolean;
  readonly e: H | undefined;
}
export interface F {
  readonly i: string;
  readonly j: q;
  readonly k: string | undefined;
  readonly l: r;
  readonly m: s | undefined;
  readonly n: t | undefined;
  readonly o: y | undefined;
  readonly p: v[] | undefined;
}

Below is in another file

import { example } from 'src/modules/ex';

exampleInfo: example | undefined;
console.log( exampleInfo.a + exampleInfo.b.i + exampleInfo.b.j + exampleInfo.b.k + exampleInfo.b.l + exampleInfo.b.o + exampleInfo.b.p + exampleInfo.c + exampleInfo.d+ exampleInfo.e)

Currently the console log is super long and easy to make mistake. Also I want to include the key(a,b,c,o,p...etc).not just the value. If I simply type

console.log(exampleInfo);

I can get the result I want (with all key and value pair). However, it's included the m&n that I don't want.

I've also tried to destruct it.

const {m,n, ...otherChar} = exampleInfo;

But it gives me error:

"Property 'm' does not exist on type 'example | undefined'.

Or

const {F.m,F.n, ... otherChar} = exampleInfo;

But still error:

 "Property 'F' does not exist on type 'example | undefined'.

Also tried this:

const {m,n, …otherChar} = exampleInfo.b;

Error changed to:

Property 'm' does not exist on type 'F | undefined'. 

How should I destruct it? or is there any other way to log them??

Answer 1

Try console.log(JSON.stringify(example))

MDN Web Docs

Edit

Sorry I didn't see the requirement clearly. Not sure this is optimal solution, but hope it helps.

let b = Object.fromEntries(Object.entries(example.b).filter(x => x[0] != 'm' && x[0] != 'n'));
const result = {...example};
result.b = b;
console.log(JSON.stringify(result));