Creating zod schema for generic interface

Question

I've got this generic interface for paginated response:

export interface PaginatedResponse<T> {
  pageIndex: number;
  pageSize: number;
  totalCount: number;
  totalPages: number;
  items: Array<T>;
}

And then I want to turn it in zod schema for runtime type checks. The approach was like this:

const PaginatedResponseSchema = z.object({
  pageIndex: z.number(),
  pageSize: z.number(),
  totalCount: z.number(),
  totalPages: z.number(),
  items: z.array(???), // <=
});

export type PaginatedResponse<T> = z.infer<typeof PaginatedResponseSchema>;

What type of array should be items in the schema?

score 7 · Accepted Answer · answered Dec 25 '22 at 14:48

7

You can create a generic function that returns a new schema given a schema for the items field. So for example you could write:

function createPaginatedResponseSchema<ItemType extends z.ZodTypeAny>(
  itemSchema: ItemType,
) {
  return z.object({
    pageIndex: z.number(),
    pageSize: z.number(),
    totalCount: z.number(),
    totalPages: z.number(),
    items: z.array(itemSchema),
  });
}

This section in the zod docs goes into more specifics about generic functions.

answered Dec 25 '22 at 14:48

Souperman

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1

Is it also possible, to then create a type, using infer, with that createSchema function? – m4chei Jul 22 '23 at 18:43
Should be `const schema = createPaginatedResponseSchema(z.string())` and then `type Schema = z.infer` – Souperman Jul 23 '23 at 02:52

Creating zod schema for generic interface

1 Answers1