Generated type validation doesn't work for non-optional any property

[tvillaren]

Bug description

When using the any keyword to accept any input for a property, like this:

export interface Citizen {
  name: any
};

ts-to-zod generates the right output

export const citizenSchema = z.object({
  name: z.any()
});

However, as mentioned in issue colinhacks/zod#1628, the type inference from zod makes the name property optional, failing the ts-to-zod generated type validation (see 52cf4f7 for a failing test case).

Indeed, when looking at the inferred type via z.infer<typeof citizenSchema> we get the following:

type Citizen = {
  name?: any
}

which doesn't match the initial type, thus fail validation.

There is no full consensus on the fix on the zod Github.
👉 An user proposed a workaround but that would mean using a custom function instead of z.object().
👉 We could ignore this specific generated type validation error, but it may come from various sources.
👉 We could live with it, but that would mean only supporting any for optional properties (and thus, warn ts-to-zod users about this upon generation).

Happy to try to fix this but not sure toward which direction we could go 🤷

[fabien0102]

Interesting! I guess we can relax the validator for this usecase. The validator is here to help discovering missing types, so I think it's fine to ignore this case, I can have a look on Monday if you want! 😊

Have a nice weekend

[tvillaren]

Hello @fabien0102,

Have you had a chance to look at this?
If not, I can give it a try but I'm not sure what to do to ignore this case. Any pointer would be nice! Thx.

[fabien0102]

As you probably guessed, I had no time whatsoever 🙈 Until now! 😁

This should be fixed, thanks again for the clear reporting and follow up