Generate declaration file with single module in TypeScript

Question

Given the following folder structure:

src/
├── foo.ts
├── bar.ts
├── baz.ts
├── index.ts

Where foo.ts, bar.ts, and baz.ts each export a default class or thing: i.e. in the case of foo.ts:

export default class Foo {
    x = 2;
}

Can we automatically generate a declaration file which declares one module my-module and exports foo.ts, bar.ts, and baz.ts as non-defaults?

I.e. I want tsc to generate the following:

build/
├── foo.js
├── bar.js
├── baz.js
├── index.js
├── index.d.ts

Where index.d.ts contains:

declare module 'my-module' {
    export class Foo {
        ...
    }
    export class Bar {
        ...
    }
    export class Baz {
        ...
    }
}

I see that mostly all NPM modules have a declaration file like this and maybe with separate files.

How would I accomplish this?

Phenylamine

This TypeScript issue is for generating a "flattened" declaration file and has links to some third-party tools that you could try. May I ask why you want this? Generating a .d.ts file per .js file will work fine for being able to find type information, though a flattened .d.ts file may be easier for users to understand and may make it easier to hide elements that you don't consider public API. – Otherdirected

Answer 1

In your tsconfig.json file, set "declaration" to true. Then when you run typescript on your project it will automatically generate declaration files for you.

Midsummer

Johan is asking how to bundle all type declaration files in one instead of emitting the type declaration file. – Autoerotic

Answer 2

tsup

It can do that and is pretty fast (1.5s):

tsup ./executors/index.ts --dts-only

You need to import the types in ./executors/index.ts and export them from this file.

dts-bundle-generator

It was slower for me (8.5s):

dts-bundle-generator --out-file dist/testkube-executors.d.ts ./executors/executors.ts

Deathlike