fs.readFileSync is not file relative? Node.js

Question

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?

__dirname is not necessarily the same as the current working directory, but fs.readFileSync() uses the current working directory. The current working directory is likely set to whatever it was when you launched node. — jfriend00
– jfriend00, Commented Jun 17, 2017 at 6:11

cartant · Accepted Answer · 2017-06-17 05:13:57Z

342

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"));

answered Jun 17, 2017 at 5:13

cartant

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4 Comments

cartant Over a year ago

Because relative paths are resolved using the current directory. And you want to resolve using a different directory. From the docs: "The relative path to a filename can be used. Remember, however, that this path will be relative to process.cwd()."

weaveoftheride Over a year ago

for some reason I got it into my head that because local file wasn't working on my remote environment that i . would need to get the file by a full url. However, of course I didn't and I was able to use this method to resolve the absolute path on my remote server..

Chris Halcrow Over a year ago

Note - path is a node module that provides utilities for working with file and directory paths. __dirname is a node.js environment variable that point to the parent directory of the current file.

Daniel Azamar Over a year ago

I will just add to take a look at the cosole... maybe some "../" is missing. It works like a charm!

Stephen Jenkins · Accepted Answer · 2021-02-18 18:29:36Z

21

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'));

answered Feb 18, 2021 at 18:29

Stephen Jenkins

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3 Comments

404 Over a year ago

Not sure I see how this expands the accepted answer, it seems exactly the same to me - path.resolve with __dirname and a relative path.

Stephen Jenkins Over a year ago

@404 Yes I was including for TypeScript folks :) the code is the same but using import statements.

riv Over a year ago

__dirname does not work in ESM, so the first example will not compile.

Renatex · Accepted Answer · 2022-09-08 14:51:08Z

5

Another alternative if you are using type module is to use process.cwd()

package.json

{
    "type": "module",
}

console.log(process.cwd() + relative_path) // /User/your_user/path_to_folder

answered Sep 8, 2022 at 14:51

Renatex

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Comments

Shiva127 · Accepted Answer · 2024-07-11 11:49:15Z

3

If you are using ES modules ("type": "module"), you can get the module path from import.meta.url:

import { fileURLToPath } from 'url'
import { dirname, resolve } from 'path'
import { readFileSync } from 'fs'

const modulePath = dirname(fileURLToPath(import.meta.url))

const file = readFileSync(resolve(modulePath, '../file.xml'))

or

import { readFileSync } from 'fs';

const file = readFileSync(new URL('../file.xml', import.meta.url));

Since Node.js version 20.11.0 you can call import.meta.dirname and import.meta.filename.

import { resolve } from 'path'
import { readFileSync } from 'fs'

const file = readFileSync(resolve(import.meta.dirname, '../file.xml'))

edited Jul 11, 2024 at 11:49

answered Oct 12, 2023 at 6:26

Shiva127

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