In a Node.js module I\'m writing I would like to open a file--i.e, with fs.readFile()
--that is contained in the same directory as my module. By which I mean it
As david van brink mentioned in the comments, the correct solution is to use __dirname
. This global variable will return the path of the currently executing script (i.e. you might need to use ../
to reach the root of your module).
For example:
var path = require("path");
require(path.join(__dirname, '/models'));
Just to save someone from a headache.