ZodRecursive types
You can define a recursive schema in Zod, but because of a limitation of TypeScript, their type can't be statically inferred. Instead you'll need to define the type definition manually, and provide it to Zod as a "type hint".
ts
const baseCategorySchema = z.object({
name: z.string(),
});
type Category = z.infer<typeof baseCategorySchema> & {
subcategories: Category[];
};
const categorySchema: z.ZodType<Category> = baseCategorySchema.extend({
subcategories: z.lazy(() => categorySchema.array()),
});
categorySchema.parse({
name: "People",
subcategories: [
{
name: "Politicians",
subcategories: [
{
name: "Presidents",
subcategories: [],
},
],
},
],
}); // passesconst baseCategorySchema = z.object({
name: z.string(),
});
type Category = z.infer<typeof baseCategorySchema> & {
subcategories: Category[];
};
const categorySchema: z.ZodType<Category> = baseCategorySchema.extend({
subcategories: z.lazy(() => categorySchema.array()),
});
categorySchema.parse({
name: "People",
subcategories: [
{
name: "Politicians",
subcategories: [
{
name: "Presidents",
subcategories: [],
},
],
},
],
}); // passesThanks to crasite for this example.
ZodType with ZodEffects
When using z.ZodType with z.ZodEffects ( .refine, .transform, preprocess, etc... ), you will need to define the input and output types of the schema. z.ZodType<Output, z.ZodTypeDef, Input>
ts
const isValidId = (id: string): id is `${string}/${string}` =>
id.split("/").length === 2;
const baseSchema = z.object({
id: z.string().refine(isValidId),
});
type Input = z.input<typeof baseSchema> & {
children: Input[];
};
type Output = z.output<typeof baseSchema> & {
children: Output[];
};
const schema: z.ZodType<Output, z.ZodTypeDef, Input> = baseSchema.extend({
children: z.lazy(() => schema.array()),
});const isValidId = (id: string): id is `${string}/${string}` =>
id.split("/").length === 2;
const baseSchema = z.object({
id: z.string().refine(isValidId),
});
type Input = z.input<typeof baseSchema> & {
children: Input[];
};
type Output = z.output<typeof baseSchema> & {
children: Output[];
};
const schema: z.ZodType<Output, z.ZodTypeDef, Input> = baseSchema.extend({
children: z.lazy(() => schema.array()),
});Thanks to marcus13371337 and JoelBeeldi for this example.
JSON type
If you want to validate any JSON value, you can use the snippet below.
ts
const literalSchema = z.union([z.string(), z.number(), z.boolean(), z.null()]);
type Literal = z.infer<typeof literalSchema>;
type Json = Literal | { [key: string]: Json } | Json[];
const jsonSchema: z.ZodType<Json> = z.lazy(() =>
z.union([literalSchema, z.array(jsonSchema), z.record(jsonSchema)])
);
jsonSchema.parse(data);const literalSchema = z.union([z.string(), z.number(), z.boolean(), z.null()]);
type Literal = z.infer<typeof literalSchema>;
type Json = Literal | { [key: string]: Json } | Json[];
const jsonSchema: z.ZodType<Json> = z.lazy(() =>
z.union([literalSchema, z.array(jsonSchema), z.record(jsonSchema)])
);
jsonSchema.parse(data);Thanks to ggoodman for suggesting this.
Cyclical objects
Despite supporting recursive schemas, passing cyclical data into Zod will cause an infinite loop.