💖 _Huge thanks to the sponsors who help me maintain this repo:_

Theodo

Your Brand Here

Stop typing twice 🙅‍♂️

A lot of projects use JSON schemas for runtime data validation along with TypeScript for static type checking.

Their code may look like this:

``typescript
const dogSchema = {
type: "object",
properties: {
name: { type: "string" },
age: { type: "integer" },
hobbies: { type: "array", items: { type: "string" } },
favoriteFood: { enum: ["pizza", "taco", "fries"] },
},
required: ["name", "age"],
};

type Dog = {
name: string;
age: number;
hobbies?: string[];
favoriteFood?: "pizza" | "taco" | "fries";
};
`

Both objects carry similar if not exactly the same information. This is a code duplication that can annoy developers and introduce bugs if not properly maintained.

That's when json-schema-to-ts comes to the rescue 💪

FromSchema

The FromSchema method lets you infer TS types directly from JSON schemas:

`typescript
import { FromSchema } from "json-schema-to-ts";

const dogSchema = {
type: "object",
properties: {
name: { type: "string" },
age: { type: "integer" },
hobbies: { type: "array", items: { type: "string" } },
favoriteFood: { enum: ["pizza", "taco", "fries"] },
},
required: ["name", "age"],
} as const;

type Dog = FromSchema;
// => Will infer the same type as above
`

Schemas can even be nested, as long as you don't forget the as const statement:

`typescript
const catSchema = { ... } as const;

const petSchema = {
anyOf: [dogSchema, catSchema],
} as const;

type Pet = FromSchema;
// => Will work 🙌
`

The as const statement is used so that TypeScript takes the schema definition to the word (e.g. _true_ is interpreted as the _true_ constant and not widened as _boolean_). It is pure TypeScript and has zero impact on the compiled code.

If you don't mind impacting the compiled code, you can use the asConst util, which simply returns the schema while narrowing its inferred type.

`typescript
import { asConst } from "json-schema-to-ts";

const dogSchema = asConst({
type: "object",
...
});

type Dog = FromSchema;
// => Will work as well 🙌
`

Since TS 4.9, you can also use the satisfies operator to benefit from type-checking and autocompletion:

`typescript
import type { JSONSchema } from "json-schema-to-ts";

const dogSchema = {
// Type-checked and autocompleted 🙌
type: "object"
...
} as const satisfies JSONSchema

type Dog = FromSchema
// => Still work 🙌
`

You can also use this with JSDocs by wrapping your schema in /* @type {const} @satisfies {import('json-schema-to-ts').JSONSchema} / (...) like:

`
const dogSchema = /* @type {const} @satisfies {import('json-schema-to-ts').JSONSchema} / ({
// Type-checked and autocompleted 🙌
type: "object"
...
})

/* @type {import('json-schema-to-ts').FromSchema} /
const dog = { ... }
`

Why use json-schema-to-ts?

If you're looking for runtime validation with added types, libraries like yup, zod or runtypes may suit your needs while being easier to use!

On the other hand, JSON schemas have the benefit of being widely used, more versatile and reusable (swaggers, APIaaS...).

If you prefer to stick to them and can define your schemas in TS instead of JSON (importing JSONs as const is not available yet), then json-schema-to-ts is made for you:

- ✅ Schema validation FromSchema raises TS errors on invalid schemas, based on DefinitelyTyped's definitions
- ✨ No impact on compiled code: json-schema-to-ts only operates in type space. And after all, what's lighter than a dev-dependency?
- 🍸 DRYness: Less code means less embarrassing typos
- 🤝 Real-time consistency: See that string that you used instead of an enum? Or this additionalProperties you confused with additionalItems? Or forgot entirely? Well, json-schema-to-ts does!
- 🔧 Reliability: FromSchema is extensively tested against AJV, and covers all the use cases that can be handled by TS for now\*
- 🏋️‍♂️ Help on complex schemas: Get complex schemas right first time with instantaneous typing feedbacks! For instance, it's not obvious the following schema can never be validated:

`typescript
const addressSchema = {
type: "object",
allOf: [
{
properties: {
street: { type: "string" },
city: { type: "string" },
state: { type: "string" },
},
required: ["street", "city", "state"],
},
{
properties: {
type: { enum: ["residential", "business"] },
},
},
],
additionalProperties: false,
} as const;
`

But it is with FromSchema!

`typescript
type Address = FromSchema;
// => never 🙌
`

> \*If json-schema-to-ts misses one of your use case, feel free to open an issue 🤗

Table of content

- Installation
- Use cases
- Const
- Enums
- Primitive types
- Nullable
- Arrays
- Tuples
- Objects
- Combining schemas
- AnyOf
- AllOf
- OneOf
- Not
- If/Then/Else
- Definitions
- References
- Deserialization
- Extensions
- Typeguards
- Validators
- Compilers
- FAQ

Installation

`bash


npm

npm install --save-dev json-schema-to-ts

yarn

yarn add --dev json-schema-to-ts
`

> json-schema-to-ts requires TypeScript 4.3+. Using strict mode is required, as well as (apparently) turning off noStrictGenericChecks.

Use cases

$3

`typescript
const fooSchema = {
const: "foo",
} as const;

type Foo = FromSchema;
// => "foo"
`

$3

`typescript
const enumSchema = {
enum: [true, 42, { foo: "bar" }],
} as const;

type Enum = FromSchema;
// => true | 42 | { foo: "bar"}
`

You can also go full circle with typescript enums.

`typescript
enum Food {
Pizza = "pizza",
Taco = "taco",
Fries = "fries",
}

const enumSchema = {
enum: Object.values(Food),
} as const;

type Enum = FromSchema;
// => Food
`

$3

`typescript
const primitiveTypeSchema = {
type: "null", // "boolean", "string", "integer", "number"
} as const;

type PrimitiveType = FromSchema;
// => null, boolean, string or number
`

`typescript
const primitiveTypesSchema = {
type: ["null", "string"],
} as const;

type PrimitiveTypes = FromSchema;
// => null | string
`

> For more complex types, refinment keywords like required or additionalItems will apply 🙌

$3

`typescript
const nullableSchema = {
type: "string",
nullable: true,
} as const;

type Nullable = FromSchema;
// => string | null
`

$3

`typescript
const arraySchema = {
type: "array",
items: { type: "string" },
} as const;

type Array = FromSchema;
// => string[]
`

$3

`typescript
const tupleSchema = {
type: "array",
items: [{ type: "boolean" }, { type: "string" }],
} as const;

type Tuple = FromSchema;
// => [] | [boolean] | [boolean, string] | [boolean, string, ...unknown[]]
`

FromSchema supports the additionalItems keyword:

`typescript
const tupleSchema = {
type: "array",
items: [{ type: "boolean" }, { type: "string" }],
additionalItems: false,
} as const;

type Tuple = FromSchema;
// => [] | [boolean] | [boolean, string]
`

`typescript
const tupleSchema = {
type: "array",
items: [{ type: "boolean" }, { type: "string" }],
additionalItems: { type: "number" },
} as const;

type Tuple = FromSchema;
// => [] | [boolean] | [boolean, string] | [boolean, string, ...number[]]
`

...as well as the minItems and maxItems keywords:

`typescript
const tupleSchema = {
type: "array",
items: [{ type: "boolean" }, { type: "string" }],
minItems: 1,
maxItems: 2,
} as const;

type Tuple = FromSchema;
// => [boolean] | [boolean, string]
`

> Additional items will only work if Typescript's strictNullChecks option is activated

$3

`typescript
const objectSchema = {
type: "object",
properties: {
foo: { type: "string" },
bar: { type: "number" },
},
required: ["foo"],
} as const;

type Object = FromSchema;
// => { [x: string]: unknown; foo: string; bar?: number; }
`

Defaulted properties (even optional ones) will be set as required in the resulting type. You can turn off this behavior by setting the keepDefaultedPropertiesOptional option to true:

`typescript
const defaultedProp = {
type: "object",
properties: {
foo: { type: "string", default: "bar" },
},
additionalProperties: false,
} as const;

type Object = FromSchema;
// => { foo: string; }

type Object = FromSchema<
typeof defaultedProp,
{ keepDefaultedPropertiesOptional: true }
>;
// => { foo?: string; }
`

FromSchema partially supports the additionalProperties, patternProperties and unevaluatedProperties keywords:

- additionalProperties and unevaluatedProperties can be used to deny additional properties.

`typescript
const closedObjectSchema = {
...objectSchema,
additionalProperties: false,
} as const;

type Object = FromSchema;
// => { foo: string; bar?: number; }
`

`typescript
const closedObjectSchema = {
type: "object",
allOf: [
{
properties: {
foo: { type: "string" },
},
required: ["foo"],
},
{
properties: {
bar: { type: "number" },
},
},
],
unevaluatedProperties: false,
} as const;

type Object = FromSchema;
// => { foo: string; bar?: number; }
`

- Used on their own, additionalProperties and/or patternProperties can be used to type unnamed properties.

`typescript
const openObjectSchema = {
type: "object",
additionalProperties: {
type: "boolean",
},
patternProperties: {
"^S": { type: "string" },
"^I": { type: "integer" },
},
} as const;

type Object = FromSchema;
// => { [x: string]: string | number | boolean }
`

However:

- When used in combination with the properties keyword, extra properties will always be typed as unknown to avoid conflicts.

`typescript
const mixedObjectSchema = {
type: "object",
properties: {
foo: { enum: ["bar", "baz"] },
},
additionalProperties: { type: "string" },
} as const;

type Object = FromSchema;
// => { [x: string]: unknown; foo?: "bar" | "baz"; }
`

- Due to its context-dependent nature, unevaluatedProperties does not type extra-properties when used on its own. Use additionalProperties instead.

`typescript
const openObjectSchema = {
type: "object",
unevaluatedProperties: {
type: "boolean",
},
} as const;

type Object = FromSchema;
// => { [x: string]: unknown }
`

Combining schemas

$3

`typescript
const anyOfSchema = {
anyOf: [
{ type: "string" },
{
type: "array",
items: { type: "string" },
},
],
} as const;

type AnyOf = FromSchema;
// => string | string[]
`

FromSchema will correctly infer factored schemas:

`typescript
const factoredSchema = {
type: "object",
properties: {
bool: { type: "boolean" },
},
required: ["bool"],
anyOf: [
{
properties: {
str: { type: "string" },
},
required: ["str"],
},
{
properties: {
num: { type: "number" },
},
},
],
} as const;

type Factored = FromSchema;
// => {
// [x:string]: unknown;
// bool: boolean;
// str: string;
// } | {
// [x:string]: unknown;
// bool: boolean;
// num?: number;
// }
`

$3

FromSchema will parse the oneOf keyword in the same way as anyOf:

`typescript
const catSchema = {
type: "object",
oneOf: [
{
properties: {
name: { type: "string" },
},
required: ["name"],
},
{
properties: {
color: { enum: ["black", "brown", "white"] },
},
},
],
} as const;

type Cat = FromSchema;
// => {
// [x: string]: unknown;
// name: string;
// } | {
// [x: string]: unknown;
// color?: "black" | "brown" | "white";
// }

// => Error will NOT be raised 😱
const invalidCat: Cat = { name: "Garfield" };
`

$3

`typescript
const addressSchema = {
type: "object",
allOf: [
{
properties: {
address: { type: "string" },
city: { type: "string" },
state: { type: "string" },
},
required: ["address", "city", "state"],
},
{
properties: {
type: { enum: ["residential", "business"] },
},
},
],
} as const;

type Address = FromSchema;
// => {
// [x: string]: unknown;
// address: string;
// city: string;
// state: string;
// type?: "residential" | "business";
// }
`

$3

Exclusions require heavy computations, that can sometimes be aborted by Typescript and end up in any inferred types. For this reason, they are not activated by default: You can opt-in with the parseNotKeyword option.

`typescript
const tupleSchema = {
type: "array",
items: [{ const: 1 }, { const: 2 }],
additionalItems: false,
not: {
const: [1],
},
} as const;

type Tuple = FromSchema;
// => [] | [1, 2]
`

`typescript
const primitiveTypeSchema = {
not: {
type: ["array", "object"],
},
} as const;

type PrimitiveType = FromSchema<
typeof primitiveTypeSchema,
{ parseNotKeyword: true }
>;
// => null | boolean | number | string
`

In objects and tuples, the exclusion will propagate to properties/items if it can collapse on a single one.

`typescript
// 👍 Can be propagated on "animal" property
const petSchema = {
type: "object",
properties: {
animal: { enum: ["cat", "dog", "boat"] },
},
not: {
properties: { animal: { const: "boat" } },
},
required: ["animal"],
additionalProperties: false,
} as const;

type Pet = FromSchema;
// => { animal: "cat" | "dog" }
`

`typescript
// ❌ Cannot be propagated
const petSchema = {
type: "object",
properties: {
animal: { enum: ["cat", "dog"] },
color: { enum: ["black", "brown", "white"] },
},
not: {
const: { animal: "cat", color: "white" },
},
required: ["animal", "color"],
additionalProperties: false,
} as const;

type Pet = FromSchema;
// => { animal: "cat" | "dog", color: "black" | "brown" | "white" }
`

As some actionable keywords are not yet parsed, exclusions that resolve to never are granted the benefit of the doubt and omitted. For the moment, FromSchema assumes that you are not crafting unvalidatable exclusions.

`typescript
const oddNumberSchema = {
type: "number",
not: { multipleOf: 2 },
} as const;

type OddNumber = FromSchema;
// => should and will resolve to "number"

const incorrectSchema = {
type: "number",
not: { bogus: "option" },
} as const;

type Incorrect = FromSchema;
// => should resolve to "never" but will still resolve to "number"
`

Also, keep in mind that TypeScript misses refinment types:

`typescript
const goodLanguageSchema = {
type: "string",
not: {
enum: ["Bummer", "Silly", "Lazy sod !"],
},
} as const;

type GoodLanguage = FromSchema<
typeof goodLanguageSchema,
{ parseNotKeyword: true }
>;
// => string
`

$3

For the same reason as the Not keyword, conditions parsing is not activated by default: You can opt-in with the parseIfThenElseKeywords option.

`typescript
const petSchema = {
type: "object",
properties: {
animal: { enum: ["cat", "dog"] },
dogBreed: { enum: Object.values(DogBreed) },
catBreed: { enum: Object.values(CatBreed) },
},
required: ["animal"],
if: {
properties: {
animal: { const: "dog" },
},
},
then: {
required: ["dogBreed"],
not: { required: ["catBreed"] },
},
else: {
required: ["catBreed"],
not: { required: ["dogBreed"] },
},
additionalProperties: false,
} as const;

type Pet = FromSchema;
// => {
// animal: "dog";
// dogBreed: DogBreed;
// catBreed?: CatBreed | undefined
// } | {
// animal: "cat";
// catBreed: CatBreed;
// dogBreed?: DogBreed | undefined
// }
`

> ☝️ FromSchema computes the resulting type as (If ∩ Then) ∪ (¬If ∩ Else). While correct in theory, remember that the not keyword is not perfectly assimilated, which may become an issue in some complex schemas.

$3

`typescript
const userSchema = {
type: "object",
properties: {
name: { $ref: "#/definitions/name" },
age: { $ref: "#/definitions/age" },
},
required: ["name", "age"],
additionalProperties: false,
definitions: {
name: { type: "string" },
age: { type: "integer" },
},
} as const;

type User = FromSchema;
// => {
// name: string;
// age: number;
// }
`

> ☝️ Wether in definitions or references, FromSchema will not work on recursive schemas for now.

$3

Unlike run-time validator classes like AJV, TS types cannot withhold internal states. Thus, they cannot keep any identified schemas in memory.

But you can hydrate them via the references option:

`typescript
const userSchema = {
$id: "http://example.com/schemas/user.json",
type: "object",
properties: {
name: { type: "string" },
age: { type: "integer" },
},
required: ["name", "age"],
additionalProperties: false,
} as const;

const usersSchema = {
type: "array",
items: {
$ref: "http://example.com/schemas/user.json",
},
} as const;

type Users = FromSchema<
typeof usersSchema,
{ references: [typeof userSchema] }
>;
// => {
// name: string;
// age: string;
// }[]

const anotherUsersSchema = {
$id: "http://example.com/schemas/users.json",
type: "array",
items: { $ref: "user.json" },
} as const;
// => Will work as well 🙌
`

Deserialization

You can specify deserialization patterns with the deserialize option:

`typescript
const userSchema = {
type: "object",
properties: {
name: { type: "string" },
email: {
type: "string",
format: "email",
},
birthDate: {
type: "string",
format: "date-time",
},
},
required: ["name", "email", "birthDate"],
additionalProperties: false,
} as const;

type Email = string & { brand: "email" };

type User = FromSchema<
typeof userSchema,
{
deserialize: [
{
pattern: {
type: "string";
format: "email";
};
output: Email;
},
{
pattern: {
type: "string";
format: "date-time";
};
output: Date;
},
];
}
>;
// => {
// name: string;
// email: Email;
// birthDate: Date;
// }
`

Extensions

If you need to extend the JSON Schema spec with custom properties, use the ExtendedJSONSchema and FromExtendedSchema types to benefit from json-schema-to-ts:

`typescript
import type { ExtendedJSONSchema, FromExtendedSchema } from "json-schema-to-ts";

type CustomProps = {
numberType: "int" | "float" | "bigInt";
};

const bigIntSchema = {
type: "number",
numberType: "bigInt",
// 👇 Ensures mySchema is correct (includes extension)
} as const satisfies ExtendedJSONSchema;

type BigInt = FromExtendedSchema<
CustomProps,
typeof bigIntSchema,
{
// 👇 Works very well with the deserialize option!
deserialize: [
{
pattern: {
type: "number";
numberType: "bigInt";
};
output: bigint;
},
];
}
>;
`

Typeguards

You can use FromSchema to implement your own typeguard:

`typescript
import { FromSchema, Validator } from "json-schema-to-ts";

// It's important to:
// - Explicitely type your validator as Validator
// - Use FromSchema as the default value of a 2nd generic first
const validate: Validator = >(
schema: S,
data: unknown
): data is T => {
const isDataValid: boolean = ... // Implement validation here
return isDataValid;
};

const petSchema = { ... } as const
let data: unknown;
if (validate(petSchema, data)) {
const { name, ... } = data; // data is typed as Pet 🙌
}
`

If needed, you can provide FromSchema options and additional validation options to the Validator type:

`typescript
type FromSchemaOptions = { parseNotKeyword: true };
type ValidationOptions = [{ fastValidate: boolean }]

const validate: Validator = <
S extends JSONSchema,
T = FromSchema
>(
schema: S,
data: unknown,
...validationOptions: ValidationOptions
): data is T => { ... };
`

json-schema-to-ts also exposes two helpers to write type guards. They don't impact the code that you wrote (they simply return it), but turn it into type guards.

You can use them to wrap either validators or compilers.

$3

A validator is a function that receives a schema plus some data, and returns true if the data is valid compared to the schema, false otherwise.

You can use the wrapValidatorAsTypeGuard helper to turn validators into type guards. Here is an implementation with ajv:

`typescript
import Ajv from "ajv";
import { $Validator, wrapValidatorAsTypeGuard } from "json-schema-to-ts";

const ajv = new Ajv();

// The initial validator definition is up to you
// ($Validator is prefixed with $ to differ from resulting type guard)
const $validate: $Validator = (schema, data) => ajv.validate(schema, data);

const validate = wrapValidatorAsTypeGuard($validate);

const petSchema = { ... } as const;

let data: unknown;
if (validate(petSchema, data)) {
const { name, ... } = data; // data is typed as Pet 🙌
}
`

If needed, you can provide FromSchema options and additional validation options as generic types:

`typescript
type FromSchemaOptions = { parseNotKeyword: true };
type ValidationOptions = [{ fastValidate: boolean }]

const $validate: $Validator = (
schema,
data,
...validationOptions // typed as ValidationOptions
) => { ... };

// validate will inherit from ValidationOptions
const validate = wrapValidatorAsTypeGuard($validate);

// with special FromSchemaOptions
// (ValidationOptions needs to be re-provided)
const validate = wrapValidatorAsTypeGuard<
FromSchemaOptions,
ValidationOptions
>($validate);
`

$3

A compiler is a function that takes a schema as an input and returns a data validator for this schema as an output.

You can use the wrapCompilerAsTypeGuard helper to turn compilers into type guard builders. Here is an implementation with ajv:

`typescript
import Ajv from "ajv";
import { $Compiler, wrapCompilerAsTypeGuard } from "json-schema-to-ts";

// The initial compiler definition is up to you
// ($Compiler is prefixed with $ to differ from resulting type guard)
const $compile: $Compiler = (schema) => ajv.compile(schema);

const compile = wrapCompilerAsTypeGuard($compile);

const petSchema = { ... } as const;

const isPet = compile(petSchema);

let data: unknown;
if (isPet(data)) {
const { name, ... } = data; // data is typed as Pet 🙌
}
`

If needed, you can provide FromSchema options, additional compiling and validation options as generic types:

`typescript
type FromSchemaOptions = { parseNotKeyword: true };
type CompilingOptions = [{ fastCompile: boolean }];
type ValidationOptions = [{ fastValidate: boolean }];

const $compile: $Compiler = (
schema,
...compilingOptions // typed as CompilingOptions
) => {
...

return (
data,
...validationOptions // typed as ValidationOptions
) => { ... };
};

// compile will inherit from all options
const compile = wrapCompilerAsTypeGuard($compile);

// with special FromSchemaOptions
// (options need to be re-provided)
const compile = wrapCompilerAsTypeGuard<
FromSchemaOptions,
CompilingOptions,
ValidationOptions
>($compile);
`

Frequently Asked Questions

- Does json-schema-to-ts work on _.json_ file schemas?
- Will json-schema-to-ts impact the performances of my IDE/compiler?
- How can I apply FromSchema on generics?
- I get a type instantiation is excessively deep and potentially infinite` error, what should I do?