Laravel Elixir：require（）在合并时不选择文件内容

So I am doing like this:

elixir(function(mix) {
    mix.sass('app.scss');
    mix.scripts('app.js');
});

app.js

require('site');

when I run gulp it produces app.js in /public with line require('site') instead of putting content of it.

Update

Based on what was suggested I did:

require('./site.js') which produced following JS

(function e(t,n,r){function s(o,u){if(!n[o]){if(!t[o]){var a=typeof require=="function"&&require;if(!u&&a)return a(o,!0);if(i)return i(o,!0);var f=new Error("Cannot find module '"+o+"'");throw f.code="MODULE_NOT_FOUND",f}var l=n[o]={exports:{}};t[o][0].call(l.exports,function(e){var n=t[o][1][e];return s(n?n:e)},l,l.exports,e,t,n,r)}return n[o].exports}var i=typeof require=="function"&&require;for(var o=0;o<r.length;o++)s(r[o]);return s})({1:[function(require,module,exports){
'use strict';

require('./site');

},{"./site":2}],2:[function(require,module,exports){
'use strict';

alert('hi');

},{}]},{},[1]);

What you need is browserify:

elixir(function(mix) {
    mix.sass('app.scss');
    mix.browserify('app.js');
});

In app.js you should get: require('./site.js');