zod-to-ts
generate TypeScript types from your Zod schema
Installation
npm install zod-to-ts zod typescript
You must be on zod@3 and typescript@4.
Usage
import { z } from 'zod'
import { zodToTs } from 'zod-to-ts'
const UserSchema = z.object({
username: z.string(),
age: z.number(),
inventory: z.object({
name: z.string(),
itemId: z.number(),
}).array(),
})
const { node } = zodToTs(UserSchema, 'User')
result:
{
username: string
age: number
inventory: {
name: string
itemId: number
}[]
}
You must pass in the identifier User
or it will default to Identifier
. This is necessary to handle cases like recursive types and native enums. zodToTs()
only returns the type value, not the actual type declaration. If you want to add an identifier to the type and create a type declaration, you can use the createTypeAlias()
utility:
import { createTypeAlias, zodToTs } from 'zod-to-ts'
const identifier = 'User'
const { node } = zodToTs(UserSchema, identifier)
const typeAlias = createTypeAlias(
node,
identifier,
)
result:
type User = {
username: string
}
zodToTs()
and createTypeAlias()
return a TS AST nodes, so if you want to get the node as a string, you can use the printNode()
utility.
zodToTs()
:
import { printNode, zodToTs } from 'zod-to-ts'
const identifier = 'User'
const { node } = zodToTs(UserSchema, identifier)
const nodeString = printNode(node)
result:
"{
username: string
age: number
inventory: {
name: string
itemId: number
}[]
}"
createTypeAlias()
:
import { createTypeAlias, printNode, zodToTs } from 'zod-to-ts'
const identifier = 'User'
const { node } = zodToTs(UserSchema, identifier)
const typeAlias = createTypeAlias(node, identifier)
const nodeString = printNode(typeAlias)
result:
"type User = {
username: string
age: number
inventory: {
name: string
itemId: number
}[]
}"
Overriding Types
You can use withGetType
to override a type, which is useful when more information is needed to determine the actual type. Unfortunately, this means working with the TS AST:
import { z } from 'zod'
import { withGetType, zodToTs } from 'zod-to-ts'
const DateSchema = withGetType(
z.instanceof(Date),
(ts) => ts.factory.createIdentifier('Date'),
)
const ItemSchema = z.object({
name: z.string(),
date: DateSchema,
})
const { node } = zodToTs(ItemSchema, 'Item')
result without withGetType
override:
type Item = {
name: string
date: any
}
result with override:
type Item = {
name: string
date: Date
}
TypeScript AST Viewer can help a lot with this if you are having trouble referencing something. It even provides copy-pastable code!
Special Cases
Lazy types default to referencing the root type (User
in the following example). It is impossible to determine what it is referencing otherwise.
import { z } from 'zod'
type User = {
username: string
friends: User[]
}
const UserSchema: z.ZodSchema<User> = z.object({
username: z.string(),
friends: z.lazy(() => UserSchema).array(),
})
const { node } = zodToTs(UserSchema, 'User')
result:
type User = {
username: string
friends: User[]
}
But what happens when the schema looks like this?
type User = {
username: string
item: {
name: string
itemId: string
}
friends: User[]
}
const friendItems = z.lazy(() => UserSchema.item).array()
const UserSchema: z.ZodSchema<User> = z.object({
username: z.string(),
item: z.object({
name: z.string(),
id: z.number(),
}),
friendItems,
})
const { node } = zodToTs(UserSchema, 'User')
result:
{
username: string
item: {
name: string
id: number
}
friendItems: User[]
}
friendItems
will still have the User
type even though it is actually referencing UserSchema["item"]
. You must provide the actual type using withGetType
:
import { z } from 'zod'
import { withGetType } from 'zod-to-ts'
type User = {
username: string
item: {
name: string
id: number
}
friends: User[]
}
const friendItems: z.Schema<User['item'][]> = withGetType(
z.lazy(() => UserSchema.item).array(),
(ts, identifier) =>
ts.factory.createIndexedAccessTypeNode(
ts.factory.createTypeReferenceNode(
ts.factory.createIdentifier(identifier),
undefined,
),
ts.factory.createLiteralTypeNode(ts.factory.createStringLiteral('item')),
),
)
const UserSchema: z.ZodSchema<User> = z.object({
username: z.string(),
item: z.object({
name: z.string(),
id: z.number(),
}),
friendItems,
})
const { node } = zodToTs(UserSchema, 'User')
result:
{
username: string
item: {
name: string
id: number
}
friendItems: User['item'][]
}
z.enum()
is always preferred, but sometimes z.nativeEnum()
is necessary. z.nativeEnum()
works similarly to z.lazy()
in that the identifier of the enum cannot be determined:
import { z } from 'zod'
import { withGetType } from 'zod-to-ts'
enum Fruit {
Apple = 'apple',
Banana = 'banana',
Cantaloupe = 'cantaloupe',
}
const fruitNativeEnum: = z.nativeEnum(
Fruit,
)
const TreeSchema = z.object({
fruit: fruitNativeEnum,
})
result:
{
fruit: unknown
}
There are two ways to solve this: provide an identifier to it or resolve all the enums inside zodToTs()
.
Option 1 - providing an identifier using withGetType()
:
import { z } from 'zod'
import { withGetType, zodToTs } from 'zod-to-ts'
enum Fruit {
Apple = 'apple',
Banana = 'banana',
Cantaloupe = 'cantaloupe',
}
const fruitNativeEnum = withGetType(
z.nativeEnum(
Fruit,
),
(ts) => ts.factory.createIdentifier('Fruit'),
)
const TreeSchema = z.object({
fruit: fruitNativeEnum,
})
const { node } = zodToTs(TreeSchema)
result:
{
fruit: Fruit
}
Option 2 - resolve enums. This is the same as before, but you just need to pass an option:
const TreeTSType = zodToTs(TreeSchema, undefined, { resolveNativeEnums: true })
result:
{
node: {
fruit: Fruit
},
store: {
nativeEnums: [
enum Fruit {
Apple = 'apple',
Banana = 'banana',
Cantaloupe = 'cantaloupe',
}
]
}
}
Note: These are not the actual values, they are TS representation. The actual values are TS AST nodes.
This option allows you to embed the enums before the schema without actually depending on an external enum type.