fs.readFileSync is not file relative? Node.js

You can resolve the path relative the location of the source file - rather than the current directory - using path.resolve:

const path = require("path");
const file = fs.readFileSync(path.resolve(__dirname, "../file.xml"));

Just to expand on the above, if you are using fs.readFileSync with TypeScript (and of course CommonJS) here's the syntax:

import fs from 'fs';
import path from 'path';

const logo = fs.readFileSync(path.resolve(__dirname, './assets/img/logo.svg'));

This is because fs.readFileSync() is resolved relative to the current working directory, see the Node.js File System docs for more info.

Source: Relative fs.readFileSync paths with Node.js

And of course, the CommonJS format:

const fs = require('fs');
const path = require('path');

const logo = fs.readFileSync(path.resolve(__dirname, './assets/img/logo.svg'));