InferType - API reference

API reference › @evolu/common › Type › InferType

type InferType<T> = T["Output"];

Defined in: packages/common/src/Type.ts:1287

Extracts the Output of a Type.

Use it to declare a named object Type Output as an interface. TypeScript then preserves the interface name in tooltips and error messages instead of expanding all of its properties.

Example

import {
  NonEmptyTrimmedString100,
  PositiveInt,
  object,
  optional,
  type InferType,
} from "@evolu/common";

const User = object({
  name: NonEmptyTrimmedString100,
  age: optional(PositiveInt),
});
interface User extends InferType<typeof User> {}