Node require absolute path

Question

I have a directory that looks like this:

-- app/
   |- models/
      |- user.js
   |- config.json

I want my user.js file to re

Answer 1

I couldn't find any better solution than to write own module

Answer 2

It seems like the module-loading functionality prepends node_modules/ to the argument. For example, place the following script into /tmp/t.js:

meow = require('meow/meow.js');

create (mkdir) the /tmp/node_modules/ and try running it with BSD's ktrace (or Linux' strace, which provides comparable functionality). Node will attempt 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.node

If you do not have the node_modules/ subdirectory next to your script, it will not lookup anywhere relative to your script at all.

Answer 3

global.__require = function (file) {
  return require(__dirname + '/' + file)
}

Place this at the top of your entry point file (e.g,. index.js). Then you can just use

const foo = __require('src/foo').

I'm using the double underscore, as an ode to __dirname but you can rename it anything!

Answer 4

I dont think we need to discuss about this problem. It is design this way for easy of developer.
If we require /config.json maybe there are alot config.json in the sub folder which will typically make it even harder to understand

-- app/
   |- controller
      |- config.json(2) 
   |- models/
      |- user.js
      |- config.json(3)
   |- config.json(1)

Now tell me how you know which one config file you need to choose.
Solution: user.js

var config1 = require('../config.json');
var config2 = require('../controller/config.json');
var config3 = require('./config.json');

if there are only one config you better put in the global variable and use it from there.

Answer 5

The simple answer is that you aren't doing anything wrong. Per a little research, the require function looks for one of:

• a core module such as fs
• a relative filepath that you specify in your require call
• a directory search for the appropriately named module in a node_modules folder somewhere in a parent directory of the file in which the require function is called.

See: http://www.bennadel.com/blog/2169-Where-Does-Node-js-And-Require-Look-For-Modules-.htm And: http://nodejs.org/api/modules.html#loading_from_node_modules_Folders

Both resources above reference the ability to also modify a NODE_PATH environment variable, however this sounds like very bad practice and at a minimum would make your code way less portable. It also probably wouldn't even work for a file like config.json since, though I'd never done it, I'd imagine changing the NODE_PATH variable only changes where require() looks for package.json files corresponding to real modules.

In summary, see: How to make the require in node.js to be always relative to the root folder of the project? as referenced above by Piotr Kowalczuk. Per that post, you have two real options:

1. Use relative file paths.
2. Package up your desired resource as a real module with a package.json file and drop it into the node_modules folder.

Finally, just bear in mind that what you are trying to do goes against the grain of the program that you are using. I think this is one case where you will ultimately make your life easier (and your collaborators'!) by going with the grain of Node and using relative file paths as the design intends.