When using require() to create a combined string path, the path of the module will not be included in the output script. Examples of this include:
require("./"+"b" );
//or
var path="./";
require(path+"b");
Let's say we have a.js
module.exports="a";
and b.js
module.exports="b";
If you use var b = require('./'+'b');, the result will be:
(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){
module.exports="a";
},{}],2:[function(require,module,exports){
var a = require('./a');
var b = require('./'+'b');
},{"./a":1}]},{},[2]);
If you use var b = require('./b');, the result will be:
(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){
module.exports="a";
},{}],2:[function(require,module,exports){
module.exports="b";
},{}],3:[function(require,module,exports){
var a = require('./a');
var b = require('./b');
},{"./a":1,"./b":2}]},{},[3]);
An issue has been opened regarding this: https://github.com/substack/node-browserify/issues/883