z.ZodSchema Questions

[nahtnam]

Hello,

I was looking for a way to define a human-readable type and then come up with the equivalent schema. This is what I ended up with:

type Dashboard = {
  id: string;
  type: string;
};
type Dashboards = {
  dashboards?: Dashboard[];
  currentDashboardId?: string;
  asdf?: string;
};

const dashboard: z.ZodSchema<Dashboard> = z.object({
  id: z.string(),
  type: z.string(),
});

export const dashboards: z.ZodSchema<Dashboards> = z.object({
  dashboards: z.array(dashboard).default([]),
  currentDashboardId: z.string().optional(),
});

Here are my two questions:

1. I noticed that here: dashboards?: Dashboard[]; I have to put a ? because there is default option provided. Is this expected behavior? I understand if it is but then it means that .default and .optional are similar and maybe the docs should note that? Any alternate solutions?
2. I noticed that any optional values such as asdf are not validated and typescript does not complain that it's not implemented

Any help would be appreciated. Thanks!

NOTE: Using infer works perfectly. I'd just prefer to have a human readable type visible to make sure that the types specified are correct.

[JacobWeisenburger]

I noticed that here: dashboards?: Dashboard[]; I have to put a ? because there is default option provided. Is this expected behavior? I understand if it is but then it means that .default and .optional are similar and maybe the docs should note that? Any alternate solutions?

If you have a required field, then giving it a default value doesn't make any sense. Because, by definition, a required field always needs to be passed into parse or it will throw an error. So defining a default value would only ever make sense if that field was optional. So the docs could be updated to be more explicit about this, but I don't think it is entirely necessary.

---

I noticed that any optional values such as asdf are not validated and typescript does not complain that it's not implemented

• This isn't a question, this is a statement. :D :P
• Because asdf is an optional property of Dashboards, then z.ZodSchema<Dashboards> doesn't care if it's implemented or not because technically it isn't required by Dashboards. So the resulting type (z.infer<typeof dashboards>) is still compatible with Dashboards. If you define asdf on the schema with an invalid value type, then typescript complains.

export const dashboards: z.ZodSchema<Dashboards> = z.object( {
//           ^^^^^^^^^^
// The types of '_type.asdf' are incompatible between these types.
// Type 'number' is not assignable to type 'string | undefined'.
    asdf: z.number(),
    dashboards: z.array( dashboard ).default( [] ),
    currentDashboardId: z.string().optional(),
} )

Hope this helps. :D