How to output only d.ts of my typescript project?

Question

For my typescript project, want to out d.ts file for entry file only:

when i have a index.ts:

import a from './a';
import b from './b';

export default {
   a,
   b
};

when i compile with ts with config tsconfig.json:

{
  "compilerOptions": {
    "outDir": "./lib",
    "declarationDir": "./build",
    "moduleResolution": "node",
    "strictNullChecks": true,
    "experimentalDecorators": true,
    "allowSyntheticDefaultImports": true,
    "noUnusedParameters": true,
    "noUnusedLocals": true,
    "noImplicitAny": true,
    "target": "es6",
    "declaration": true,
    "lib": [
      "dom",
      "es2015",
      "es2016",
    ],
  },
  "include": [
    "packages/index.ts",
    "typings"
  ],
  "exclude": [
    "node_modules",
    "**/__tests__",
    "**/__mocks__",
  ]
}

would output tree file a.d.ts, b.d.ts, index.d.ts

how only output index.d.ts

Answer 1

You can use a separate tsc run with outFile to produce a single .d.ts file that has a declare module block for each original module. You could do this by removing the declaration and declarationDir options from your original tsconfig.json and creating a separate tsconfig.declaration.json with:

{
  "extends": "./tsconfig.json",
  "compilerOptions": {
    "module": "amd",  // https://github.com/Microsoft/TypeScript/issues/27117
    "declaration": true,
    "emitDeclarationOnly": true,
    "outFile": "build/index.d.ts"
  }
}

The only problem is that TypeScript 3.0.3 incorrectly passes through relative import paths such as ./a to the output, so for now, you'll need to set up your module resolution options so that you can use non-relative import paths. The problem has been fixed in this pull request, but the fix has not been released as of 2018-09-15.

See this long thread for discussion of other approaches to declaration bundling.