fs.readFileSync is not file relative? Node.js

-Suppose I have a file at the root of my project called file.xml.

-Suppose I have a test file in tests/ called "test.js" and it has

const file = fs.readFileSync("../file.xml");

If I now run node ./tests/test.js from the root of my project it says ../file.xml does not exist. If I run the same command from within the tests directory, then it works.

It seems fs.readFileSync is relative to the directory where the script is invoked from, instead of where the script actually is. If I wrote fs.readFileSync("./file.xml") in test.js it would look more confusing and is not consistent with relative paths in a require statement which are file relative.

Why is this? How can I avoid having to rewrite the paths in my fs.readFileSync?

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'));