node.js require all files in a folder?

I recommend using glob to accomplish that task.

var glob = require( 'glob' )
  , path = require( 'path' );

glob.sync( './routes/**/*.js' ).forEach( function( file ) {
  require( path.resolve( file ) );
});

Base on @tbranyen's solution, I create an index.js file that load arbitrary javascripts under current folder as part of the exports.

// Load `*.js` under current directory as properties
//  i.e., `User.js` will become `exports['User']` or `exports.User`
require('fs').readdirSync(__dirname + '/').forEach(function(file) {
  if (file.match(/\.js$/) !== null && file !== 'index.js') {
    var name = file.replace('.js', '');
    exports[name] = require('./' + file);
  }
});

Then you can require this directory from any where else.

When require is given the path of a folder, it'll look for an index.js file in that folder; if there is one, it uses that, and if there isn't, it fails.

It would probably make most sense (if you have control over the folder) to create an index.js file and then assign all the "modules" and then simply require that.

yourfile.js

var routes = require("./routes");

index.js

exports.something = require("./routes/something.js");
exports.others = require("./routes/others.js");

If you don't know the filenames you should write some kind of loader.

Working example of a loader:

var normalizedPath = require("path").join(__dirname, "routes");

require("fs").readdirSync(normalizedPath).forEach(function(file) {
  require("./routes/" + file);
});

// Continue application logic here