Node.js + Webpack + TypeScript: access to path of project with source files (not usage project)

Question

Let's make clear once again: I don't need process.cwd in this question, I need to access to absolute path of source project. Eg:

• Source code: C:\\Users\\user1\\projects\\lib1\\src\\library.ts (becomes to Node Module in the future)
• Project that uses Library: C:\\Users\\user1\\projects\\someProject\\src\\someProject.ts

So, I need to get the C:\\Users\\user1\\projects\\lib1\\src inside library.ts .

I tried:

webpack.config.js

module.exports = {

  // ...

  target: 'node',
  externals: [nodeExternals()],

  plugins: [
    new Webpack.DefinePlugin({
      __PROJECT_ROUTE_ABSOLUTE_PATH__: __dirname
    })
  ]
};

project-types.d.ts

declare var __PROJECT_ROUTE_ABSOLUTE_PATH__: string;

If to try console.log(__PROJECT_ROUTE_ABSOLUTE_PATH__) in library.ts , below invalid JavaScript will be produced:

console.log(C:\Users\user1\projects\lib1);

The path is correct, but quotations are missing. I don't know how to explain it. But anyway, how we can get right path?

There is also a strange phenomena: if to invoke __dirname , just / will be returned, so path.resolve(__dirname, 'fileName') gives C:\\fileName `

Answer 1

You can directly use the node.js path module which is in built. The path module provides utilities for working with file and directory paths. It can be accessed using:

const path = require('path');

__filename is the file name of the current module. This is the resolved absolute path of the current module file. (ex:/home/user/some/dir/file.js)

__dirname is the directory name of the current module. (ex:/home/user/some/dir)

fs.readFile(path.resolve(__dirname + 'fileName'))

This will resolve to the path of the file.