如何在不保留示例对象的前提下生成其TypeScript类型？

Totally get this frustration—you want to leverage TypeScript’s ability to infer types from real-looking example data, but you don’t want that dummy object cluttering up your compiled JavaScript. Here are a couple of straightforward fixes:

Solution 1: Inline the Example in the Type Alias #

Instead of declaring a separate runtime variable, stick your example object directly inside the typeof operator when defining your type. Since this is purely a type-level operation, TypeScript won’t emit any of this code to your final JS bundle:

type SuccessResponse = typeof ({
  rates: [
    { cost: 10, description: "10$ shipping" },
    { cost: 20, description: "20$ shipping", guaranteedDaysToDelivery: 5 }
  ]
} as const);

The as const modifier is important here—it tells TypeScript to infer the most specific types possible (like literal values for cost instead of a generic number, and exact string literals for description), which is probably what you want.

Solution 2: Move the Example to a Declaration File #

If you prefer to keep your example data organized separately, pop it into a .d.ts file. Declaration files are never compiled to JavaScript, so your example stays purely a source for type inference:

1. Create a file like api-types.d.ts in your project:

const exampleResponse = {
  rates: [
    { cost: 10, description: "10$ shipping" },
    { cost: 20, description: "20$ shipping", guaranteedDaysToDelivery: 5 }
  ]
} as const;

export type SuccessResponse = typeof exampleResponse;

2. Import the type into your main TypeScript files whenever you need it:

import type { SuccessResponse } from './api-types';

Either of these approaches will let you keep the inferred type without the extra runtime baggage. Pick whichever fits better with how you organize your code!

内容的提问来源于stack exchange，提问作者mikeysee