I want to import a dependency of a dependency. For example, I want to import jade-load directly into my app:
my-app
┗━jade
┗━jade-load
I could do require('jade/node_modules/jade-load'), but this won't work if the node_modules tree has been flattened or deduped.
I thought of using require.resolve() to find out where jade-load really is, but there doesn't seem to be a way to tell it the starting point for the resolution. I need to be able to say "require jade-load from wherever jade is".
NB. I do not want to install jade-load as a direct dependency of my app; the point is to import the same instance that jade uses, so I can monkeypatch it.
I guess you may want to use proxyquire for managing dependencies of required modules. You can set proxyquire to globally override methods of the submodule when it will be loaded.
main.js
var proxyquire = require('proxyquire');
// use default childModule properties and methods unless they are redefined here
var childModuleStub = {
/* redefine here some necessary methods */
'#global': true
};
// parent module itself does't require childModule
var parentModuleStubs = {
'./childModule': childModuleStub
};
var parentModule = proxyquire('./parentModule', parentModuleStubs);
var result;
result = parentModule.exec();
console.log(result);
childModuleStub.data.sentence = "Overridden property.";
result = parentModule.exec();
console.log(result);
childModuleStub._exec = function () {
return "Overridden function.";
};
result = parentModule.exec();
console.log(result);
parentModule.js
var intermediateLibrary = require('./intermediateLibrary');
module.exports = {
exec: function() {
return intermediateLibrary.exec();
}
};
intermediateLibrary.js
var childModule = require('./childModule');
module.exports = {
exec: function() {
return childModule._exec();
}
};
childModule.js
var lib = {};
lib.data = {};
lib.data.sentence = "Hello, World!";
lib._exec = function () {
return lib.data.sentence;
};
module.exports = lib;
Results:
Hello, World!
Overridden property.
Overridden function.
I created the following 3 files:
base.js
var base = {};
base.one = 1;
base.two = 2;
base.three = 3;
base.bar = function(){
console.log( this.three );
};
a.js
var base = require('./base');
base.three = 6;
module.exports = base;
b.js
var base = require('./base');
module.exports = base;
test.js
var test_modules = ['a','b'];
test_modules.forEach( function( module_name ){
require( './' + module_name ).bar();
});
And then run test.js like so:
node ./test.js
It outputs this:
6
6
Why is it that when I set the property 'three' of module 'base' in 'a.js', it then affects the object in 'b.js'?
When you require() a module, it is evaluated once and cached so that subsequent require()s for the same module do not have to get loaded from disk and thus get the same exported object. So when you mutate exported properties, all references to that module will see the updated value.
You are introducing global state for base module.
The module a mutated base and then exported it as well, which means that any further references to base will have an updated value.
It is best demonstrated by the following script inside test.js
var testModules = ['b', 'a'];
testModules.forEach(function(module) {
require('./' + module).bar();
});
Now when you run node test.js, you'll see
3
6
Why?
Because the order of inclusion of the modules changed.
How do I solve this?
Simple, get rid of global state. One option is to use prototypes like so
var Base = function() {
this.one = 1;
this.two = 2;
this.three = 3;
};
Base.prototype.bar = function() {
console.log(this.three);
};
module.exports = Base;
And then, inside a.js
var Base = require('./base');
var baseInstance = new Base();
baseInstance.three = 6;
module.exports = baseInstance;
And inside b.js
var Base = require('./base');
module.exports = new Base();
Now when you run your original test.js, the output should be
6
3
I have multiple modules and I did:
// module1.js
module.exports = function() {
...
}
// module2.js
module.exports = function() {
...
}
in app.js
m1 = require('./module1')
m2 = require('./module2')
m1.method()
m2.method()
I get TypeError. Then I ended up exporting methods in both the modules.
is there a way I can export multiple modules other than exporting individual methods explicitly?
It looks like you're attempting to pass require() an undefined variable two times. require() needs to take a string as an argument to determine what module you want to load.
If the other two modules are in the same directory as app.js, try
m1 = require('./module1')
m2 = require('./module2')
EDIT:
What you're forgetting to do is
m1 = new require('./module1')()
m2 = new require('./module2')()
Assuming that you modules look like:
module.exports = function() {
this.method = function(){}
}
Personally, instead of a function I would just return an object literal from my module:
module.exports = {
method1: function(){},
method2: function(){}
}
Then I could invoke the methods from the module's export like such:
m1 = require('./module1');
m1.method1();
I'm trying to get the code below under test when occurred to me that I already included express at the top of this file. Can you some how monkey patch the express object after it's already loaded?
var express = require('express')
Helper = (function() {
var HelperObject = function(params) {
this.directories = params.directories;
};
HelperObject.prototype.addStaticPath = function(app) {
for(i = 0; i < this.directories.length; i++) {
var static = express.static('/public');
app.use(static);
}
};
return HelperObject;
})();
The problem is that when you create a node module the required modul is bound in the closure of the module and you can't start spying on it cause it isn't visible in your test.
There is Gently where you can override require but it will sprinkle your code with boilerplate test related code.
From the docs:
Returns a new require functions that catches a reference to all
required modules into gently.hijacked.
To use this function, include a line like this in your 'my-module.js'.
if (global.GENTLY) require = GENTLY.hijack(require);
var sys = require('sys');
exports.hello = function() {
sys.log('world');
};
Now you can write a test for the module above:
var gently = global.GENTLY = new (require('gently'))
, myModule = require('./my-module');
gently.expect(gently.hijacked.sys, 'log', function(str) {
assert.equal(str, 'world');
});
myModule.hello();
How do I require all files in a folder in node.js?
need something like:
files.forEach(function (v,k){
// require routes
require('./routes/'+v);
}};
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
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.
Another option is to use the package require-dir which let's you do the following. It supports recursion as well.
var requireDir = require('require-dir');
var dir = requireDir('./path/to/dir');
I have a folder /fields full of files with a single class each, ex:
fields/Text.js -> Test class
fields/Checkbox.js -> Checkbox class
Drop this in fields/index.js to export each class:
var collectExports, fs, path,
__hasProp = {}.hasOwnProperty;
fs = require('fs');
path = require('path');
collectExports = function(file) {
var func, include, _results;
if (path.extname(file) === '.js' && file !== 'index.js') {
include = require('./' + file);
_results = [];
for (func in include) {
if (!__hasProp.call(include, func)) continue;
_results.push(exports[func] = include[func]);
}
return _results;
}
};
fs.readdirSync('./fields/').forEach(collectExports);
This makes the modules act more like they would in Python:
var text = new Fields.Text()
var checkbox = new Fields.Checkbox()
One more option is require-dir-all combining features from most popular packages.
Most popular require-dir does not have options to filter the files/dirs and does not have map function (see below), but uses small trick to find module's current path.
Second by popularity require-all has regexp filtering and preprocessing, but lacks relative path, so you need to use __dirname (this has pros and contras) like:
var libs = require('require-all')(__dirname + '/lib');
Mentioned here require-index is quite minimalistic.
With map you may do some preprocessing, like create objects and pass config values (assuming modules below exports constructors):
// Store config for each module in config object properties
// with property names corresponding to module names
var config = {
module1: { value: 'config1' },
module2: { value: 'config2' }
};
// Require all files in modules subdirectory
var modules = require('require-dir-all')(
'modules', // Directory to require
{ // Options
// function to be post-processed over exported object for each require'd module
map: function(reqModule) {
// create new object with corresponding config passed to constructor
reqModule.exports = new reqModule.exports( config[reqModule.name] );
}
}
);
// Now `modules` object holds not exported constructors,
// but objects constructed using values provided in `config`.
I know this question is 5+ years old, and the given answers are good, but I wanted something a bit more powerful for express, so i created the express-map2 package for npm. I was going to name it simply express-map, however the people at yahoo already have a package with that name, so i had to rename my package.
1. basic usage:
app.js (or whatever you call it)
var app = require('express'); // 1. include express
app.set('controllers',__dirname+'/controllers/');// 2. set path to your controllers.
require('express-map2')(app); // 3. patch map() into express
app.map({
'GET /':'test',
'GET /foo':'middleware.foo,test',
'GET /bar':'middleware.bar,test'// seperate your handlers with a comma.
});
controller usage:
//single function
module.exports = function(req,res){
};
//export an object with multiple functions.
module.exports = {
foo: function(req,res){
},
bar: function(req,res){
}
};
2. advanced usage, with prefixes:
app.map('/api/v1/books',{
'GET /': 'books.list', // GET /api/v1/books
'GET /:id': 'books.loadOne', // GET /api/v1/books/5
'DELETE /:id': 'books.delete', // DELETE /api/v1/books/5
'PUT /:id': 'books.update', // PUT /api/v1/books/5
'POST /': 'books.create' // POST /api/v1/books
});
As you can see, this saves a ton of time and makes the routing of your application dead simple to write, maintain, and understand. it supports all of the http verbs that express supports, as well as the special .all() method.
npm package: https://www.npmjs.com/package/express-map2
github repo: https://github.com/r3wt/express-map
Expanding on this glob solution. Do this if you want to import all modules from a directory into index.js and then import that index.js in another part of the application. Note that template literals aren't supported by the highlighting engine used by stackoverflow so the code might look strange here.
const glob = require("glob");
let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
/* see note about this in example below */
allOfThem = { ...allOfThem, ...require(file) };
});
module.exports = allOfThem;
Full Example
Directory structure
globExample/example.js
globExample/foobars/index.js
globExample/foobars/unexpected.js
globExample/foobars/barit.js
globExample/foobars/fooit.js
globExample/example.js
const { foo, bar, keepit } = require('./foobars/index');
const longStyle = require('./foobars/index');
console.log(foo()); // foo ran
console.log(bar()); // bar ran
console.log(keepit()); // keepit ran unexpected
console.log(longStyle.foo()); // foo ran
console.log(longStyle.bar()); // bar ran
console.log(longStyle.keepit()); // keepit ran unexpected
globExample/foobars/index.js
const glob = require("glob");
/*
Note the following style also works with multiple exports per file (barit.js example)
but will overwrite if you have 2 exports with the same
name (unexpected.js and barit.js have a keepit function) in the files being imported. As a result, this method is best used when
your exporting one module per file and use the filename to easily identify what is in it.
Also Note: This ignores itself (index.js) by default to prevent infinite loop.
*/
let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
allOfThem = { ...allOfThem, ...require(file) };
});
module.exports = allOfThem;
globExample/foobars/unexpected.js
exports.keepit = () => 'keepit ran unexpected';
globExample/foobars/barit.js
exports.bar = () => 'bar run';
exports.keepit = () => 'keepit ran';
globExample/foobars/fooit.js
exports.foo = () => 'foo ran';
From inside project with glob installed, run node example.js
$ node example.js
foo ran
bar run
keepit ran unexpected
foo ran
bar run
keepit ran unexpected
One module that I have been using for this exact use case is require-all.
It recursively requires all files in a given directory and its sub directories as long they don't match the excludeDirs property.
It also allows specifying a file filter and how to derive the keys of the returned hash from the filenames.
Require all files from routes folder and apply as middleware. No external modules needed.
// require
const { readdirSync } = require("fs");
// apply as middleware
readdirSync("./routes").map((r) => app.use("/api", require("./routes/" + r)));
I'm using node modules copy-to module to create a single file to require all the files in our NodeJS-based system.
The code for our utility file looks like this:
/**
* Module dependencies.
*/
var copy = require('copy-to');
copy(require('./module1'))
.and(require('./module2'))
.and(require('./module3'))
.to(module.exports);
In all of the files, most functions are written as exports, like so:
exports.function1 = function () { // function contents };
exports.function2 = function () { // function contents };
exports.function3 = function () { // function contents };
So, then to use any function from a file, you just call:
var utility = require('./utility');
var response = utility.function2(); // or whatever the name of the function is
Can use : https://www.npmjs.com/package/require-file-directory
Require selected files with name only or all files.
No need of absoulute path.
Easy to understand and use.
Using this function you can require a whole dir.
const GetAllModules = ( dirname ) => {
if ( dirname ) {
let dirItems = require( "fs" ).readdirSync( dirname );
return dirItems.reduce( ( acc, value, index ) => {
if ( PATH.extname( value ) == ".js" && value.toLowerCase() != "index.js" ) {
let moduleName = value.replace( /.js/g, '' );
acc[ moduleName ] = require( `${dirname}/${moduleName}` );
}
return acc;
}, {} );
}
}
// calling this function.
let dirModules = GetAllModules(__dirname);
Create an index.js file in your folder with this code :
const fs = require('fs')
const files = fs.readdirSync('./routes')
for (const file of files) {
require('./'+file)
}
And after that you can simply load all the folder with require("./routes")
If you include all files of *.js in directory example ("app/lib/*.js"):
In directory app/lib
example.js:
module.exports = function (example) { }
example-2.js:
module.exports = function (example2) { }
In directory app create index.js
index.js:
module.exports = require('./app/lib');