My directory structure is as follows:
-- app/
|- models/
|- user.js
|- config.json
I am trying to make my user.js file require the config.json. Currently, I am using require('/config') but it doesn't seem to be working. Can someone please help me figure out what I am doing wrong? I would prefer not to use require('../config').
It's important to note that there is no right or wrong way when working with the require function in Node.js. Through some research, it has been discovered that the require function searches for a core module like fs, a specified relative filepath, or looks for modules in a node_modules folder within a parent directory.
While it is possible to modify the NODE_PATH environment variable, this practice is not recommended as it can lead to code portability issues. It may not work effectively for files like config.json either, as changing the NODE_PATH typically affects where require() looks for package.json files of actual modules.
For better guidance on handling require paths in node.js, refer to this post mentioned by Piotr Kowalczuk. The two main options suggested are using relative file paths or packaging resources as real modules with a package.json file placed in the node_modules folder.
Lastly, it's essential to align your approach with Node.js conventions for smoother development. Embracing relative file paths as intended by the design will likely simplify your workflow and enhance collaboration among team members.
When I couldn't find a suitable solution, I took matters into my own hands and created my own module.
After successfully testing it in various projects, I made the decision to release it as an open-source project.
const user = require('../../../database/user'); // 👎 the old way
// OR
const user = require('$db/user'); // 👍 works regardless of directory depth
const product = require('/database/product'); // 👍 easy aliasing or root path reference
It's simple to get started with just three steps:
Begin by installing the package: npm install sexy-require --save
Add require('sexy-require') at the beginning of your main application file.
require('sexy-require');
const routers = require('/routers');
const api = require('$api');
...
For advanced users, custom path configurations can be set in a .paths file located at the root of your project.
$db = /server/database
$api-v1 = /server/api/legacy
$api-v2 = /server/api/v2
Enjoy the convenience of using shortcut paths anywhere in your project:
const path = require(`sexy-require`);
console.log(path.$db); // -> '/full/path/to/app/server/database'
It appears that the module-loading feature adds node_modules/ to the argument. For instance, place the script below in /tmp/t.js:
meow = require('meow/meow.js');
create (mkdir) the /tmp/node_modules/ and test it out with BSD's ktrace (or Linux' strace, which offers similar functionality). Node will try to open the following:
/tmp/node_modules/meow/meow.js/tmp/node_modules/meow/meow.js.js/tmp/node_modules/meow/meow.js.json/tmp/node_modules/meow/meow.js.node/tmp/node_modules/meow/meow.js/package.json/tmp/node_modules/meow/meow.js/index.js/tmp/node_modules/meow/meow.js/index.json/tmp/node_modules/meow/meow.js/index.nodeIf you don't have the node_modules/ subdirectory next to your script, it won't search relative to your script at all.
function importFile(file) {
return require(__dirname + '/' + file)
}
To integrate this function into your main file, simply add it at the beginning (e.g., app.js). Then you can easily call
const bar = importFile('utils/bar')The use of double underscores is reminiscent of __dirname, but feel free to rename the function as needed!
There's no need for us to delve into this problem further. It was intentionally designed this way to make it easier for developers.
If we were to require /config.json, there could potentially be numerous config.json files in subfolders, which would only serve to complicate matters.
-- app/
|- controller
|- config.json(2)
|- models/
|- user.js
|- config.json(3)
|- config.json(1)
So, how can you determine which config file to use?
Here's the solution:
user.js
var config1 = require('../config.json');
var config2 = require('../controller/config.json');
var config3 = require('./config.json');
If there's only one configuration file, it's best to store it in a global variable and reference it from there.
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