Suggestions for crafting a type in typescript that ensures certain properties are excluded when employing the spread operator?

Question

Suggestions for crafting a type in typescript that ensures certain properties are excluded when employing the spread operator?

I have a situation where I need to send a backend model to the frontend with certain properties excluded, and I want to use typing to ensure these exclusions.

My initial attempt involved creating an interface for the model and then a restricted interface using Omit<> to exclude specific properties.

interface Thing {
    foo: string
    secret: string
}

interface RestrictedThing extends Omit<Base, "secret"> {}

const thing: Thing = { foo: "test", secret: "must_exclude_in_restricted"}

const restricted: RestrictedThing = { ...thing } //no warning is raised

However, TypeScript does not give a warning when using the spread operator to copy properties, even if they should not be part of RestrictedThing.

I came across ts-essentials StrictOmit, but it only validates the properties passed in, making it unsuitable for my needs.

I also attempted to use typeof but without success.

declare var { bar, ...forRestricted }: Thing;
type RestrictedThing = typeof forRestricted;

Is there a reliable method to create a copy of my Thing object that uses typing to exclude unwanted properties?

Any alternative approaches would be appreciated.

typescript spread-syntax

Answer 1

Answer №1

It's unfortunate that TypeScript's type system does not currently offer support for excess property checks on object literals that result from a spread operation. This limitation is well-known in the TypeScript community.

If you want to make sure that an object does not contain certain properties, you will have to manually verify those properties at runtime.

Utilize the is operator to confirm the type correctness and, within that function, remove the sensitive keys.

interface Thing {
  foo: string
  secret: string;
  secret2: string;
}
const restrictedKeys = ['secret','secret2'] as const;
interface RestrictedThing extends Omit<Thing, typeof restrictedKeys[number]> {}

const thing = {
  foo: "test",
  secret: "must_exclude_in_restricted"
};

const restricted: Partial<Thing> = { ...thing };

const cleanRestricted = (obj: Partial<Thing>): obj is RestrictedThing => {
  restrictedKeys.forEach(k => delete obj[k])
  return true;
};
if (cleanRestricted(restricted)) {
  // restricted is now really <RestrictedThing>
  restricted.secret = 'secret'; //throws error
}

Answer 2

It's unfortunate that TypeScript's type system does not currently offer support for excess property checks on object literals that result from a spread operation. This limitation is well-known in the TypeScript community.

If you want to make sure that an object does not contain certain properties, you will have to manually verify those properties at runtime.

Utilize the is operator to confirm the type correctness and, within that function, remove the sensitive keys.

interface Thing {
  foo: string
  secret: string;
  secret2: string;
}
const restrictedKeys = ['secret','secret2'] as const;
interface RestrictedThing extends Omit<Thing, typeof restrictedKeys[number]> {}

const thing = {
  foo: "test",
  secret: "must_exclude_in_restricted"
};

const restricted: Partial<Thing> = { ...thing };

const cleanRestricted = (obj: Partial<Thing>): obj is RestrictedThing => {
  restrictedKeys.forEach(k => delete obj[k])
  return true;
};
if (cleanRestricted(restricted)) {
  // restricted is now really <RestrictedThing>
  restricted.secret = 'secret'; //throws error
}

Suggestions for crafting a type in typescript that ensures certain properties are excluded when employing the spread operator?

Answer №1

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