pre-load / pre-require directories of .js route files - node.js

Using Express with Node.js, we might do something like this:
app.use('api/:controller/:action/:id', function(req,res,next){
var controller = req.params.controller;
var action = req.params.action;
var route = require('./routes/' + controller + '/' + action);
route(req,res,next);
}
now this is all fine and well, except there is at least one problem: the route file is dynamically loaded at runtime if this file has not been 'require'd yet. Which means it's a little bit slower at least.
Does someone have a script that recurses through a directory and pre-loads/pre-requires all the .js files when a server first starts up?
I have a similar problem for the front-end as well, using RequireJS. The solution seems to be to write a bash script that writes out all the .js filepaths in a directory and its subdirectories to a text file. then when the server starts up, it reads that text file and requires all the files in the directory that are listed in the text file. Is that the best way to do it?

If you can use io.js, it can preload modules using command-line -r or --require:
iojs -r <module_name> server.js

I created an NPM module that does this for the front-end, doing it for Node.js / CommonJS is another story.
https://www.npmjs.com/package/requirejs-metagen
you can use it like so:
var grm = require('requirejs-metagen'); //you can use with Gulp
var controllersOpts = {
inputFolder: './public/static/app/js/controllers/all',
appendThisToDependencies: 'app/js/controllers/',
appendThisToReturnedItems: '',
eliminateSharedFolder: true,
output: './public/static/app/js/meta/allControllers.js'
};
grm(controllersOpts,function(err){
//handle errors your own way
});
it generates a corresponding AMD/RequireJS module like so:
define(
[
"app/js/controllers/all/jobs",
"app/js/controllers/all/users"
],
function(){
return {
"jobs": arguments[0],
"users": arguments[1]
}
});
you can also require subdirectories and all that stuff like so:
var allViewsOpts = {
inputFolder: './public/static/app/js/jsx',
appendThisToDependencies: 'app/js/',
appendThisToReturnedItems: '',
eliminateSharedFolder: true,
output: './public/static/app/js/meta/allViews.js'
}
grm(allViewsOpts );
which generates output like so:
define([
"app/js/jsx/BaseView",
"app/js/jsx/reactComponents/FluxCart",
"app/js/jsx/reactComponents/FluxCartApp",
"app/js/jsx/reactComponents/FluxProduct",
"app/js/jsx/reactComponents/Item",
"app/js/jsx/reactComponents/Job",
"app/js/jsx/reactComponents/JobsList",
"app/js/jsx/reactComponents/listView",
"app/js/jsx/reactComponents/Picture",
"app/js/jsx/reactComponents/PictureList",
"app/js/jsx/reactComponents/RealTimeSearchView",
"app/js/jsx/reactComponents/Service",
"app/js/jsx/reactComponents/ServiceChooser",
"app/js/jsx/reactComponents/todoList",
"app/js/jsx/relViews/getAll/getAll",
"app/js/jsx/relViews/jobs/jobsView",
"app/js/jsx/standardViews/dashboardView",
"app/js/jsx/standardViews/overviewView",
"app/js/jsx/standardViews/pictureView",
"app/js/jsx/standardViews/portalView",
"app/js/jsx/standardViews/registeredUsersView",
"app/js/jsx/standardViews/userProfileView"
],
function(){
return {
"BaseView": arguments[0],
"reactComponents/FluxCart": arguments[1],
"reactComponents/FluxCartApp": arguments[2],
"reactComponents/FluxProduct": arguments[3],
"reactComponents/Item": arguments[4],
"reactComponents/Job": arguments[5],
"reactComponents/JobsList": arguments[6],
"reactComponents/listView": arguments[7],
"reactComponents/Picture": arguments[9],
"reactComponents/PictureList": arguments[10],
"reactComponents/RealTimeSearchView": arguments[11],
"reactComponents/Service": arguments[12],
"reactComponents/ServiceChooser": arguments[13],
"relViews/getAll/getAll": arguments[14],
"relViews/jobs/jobsView": arguments[15],
"standardViews/dashboardView": arguments[16],
"standardViews/overviewView": arguments[17],
"standardViews/pictureView": arguments[18],
"standardViews/portalView": arguments[19],
"standardViews/registeredUsersView": arguments[20],
"standardViews/userProfileView": arguments[21]
}
});
I need to update the library so it returns the stream so you can handle when it completes, otherwise it works great.

Related

fs.writeFileSync function doesn't write to file when included as a module

Consider the following:
conversations.json : []
db.js :
let fs = require('fs');
let conversations = require('./conversations.json');
function addConversation(conversation){
console.log(conversations);
conversations.push(conversation);
try{
fs.writeFileSync('conversations.json', JSON.stringify(conversations));
}
catch(err){
console.error('Parse/WriteFile Error', err)
}
}
module.exports = {
addConversation
}
app.js :
let database = require('./db.js');
database.addConversation(
{
key1: '1233',
key2: '433',
key3: '33211'
}
);
Running:
node app.js
No error is being raised. Everything compiled as expected. The problem is that the conversations.json isn't being updated once the addConversation function is called from app.js.
What's interesting is that once the addConversation is called within the db.js everything works great and the conversations.json is being updated.
What am I missing?
What am I missing?
Probably when loading as a module, you're writing the file to the wrong directory.
When you do this:
fs.writeFileSync('conversations.json', JSON.stringify(conversations));
That writes conversations.json to the current working directory which may or may not be your module directory. If you want it written to your module directory which is where this:
let conversations = require('./conversations.json');
will read it from, then you need to use __dirname to manufacture the appropriate path.
fs.writeFileSync(path.join(__dirname, 'conversations.json'), JSON.stringify(conversations));
require() automatically looks in the current module's directory when you use ./filename, but fs.writeFileSync() uses the current working directory, not your module's directory.

Gulp, livereload, jade

Need help.
I use gulp-conect and it livereload method. But if I build a few template in time, get a lot of page refresh. Is any solution, I want to build few templates with single page refresh?
So, I reproduce the problem you have and came accross this working solution.
First, lets check gulp plugins you need:
gulp-jade
gulp-livereload
optional: gulp-load-plugins
In case you need some of them go to:
http://gulpjs.com/plugins/
Search for them and install them.
Strategy: I created a gulp task called live that will check your *.jade files, and as you are working on a certain file & saving it, gulp will compile it into html and refresh the browser.
In order to accomplish that, we define a function called compileAndRefresh that will take the file returned by the watcher. It will compile that file into html and the refesh the browser (test with livereload plugin for chrome).
Notes:
I always use gulp-load-plugin to load plugins, so thats whay I use plugins.jad and plugins.livereload.
This will only compile files that are saved and while you have the task live exucting on the command line. Will not compile other files that are not in use. In order to accomplish that, you need to define a task that compiles all files, not only the ones that have been changed.
Assume .jade files in /jade and html output to /html
So, here is the gulpfile.js:
var gulp = require('gulp'),
gulpLoadPlugins = require('gulp-load-plugins'),
plugins = gulpLoadPlugins();
gulp.task('webserver', function() {
gulp.src('./html')
.pipe(plugins.webserver({
livereload: true
}));
gulp.watch('./jade/*.jade', function(event) {
compileAndRefresh(event.path);
});
});
function compileAndRefresh(file) {
gulp.src(file)
.pipe(plugins.jade({
}))
.pipe(gulp.dest('./html'))
}
Post edit notes:
Removed liveReload call from compileAndRefresh (webserver will do that).
Use gulp-server plugin insted of gulp-connect, as they suggest on their repository: "New plugin based on connect 3 using the gulp.src() API. Written in plain javascript. https://github.com/schickling/gulp-webserver"
Something you can do is to watch only files that changes, and then apply a function only to those files that have been changed, something like this:
gulp.task('live', function() {
gulp.watch('templates/folder', function(event) {
refresh_templates(event.path);
});
});
function refresh_templates(file) {
return
gulp.src(file)
.pipe(plugins.embedlr())
.pipe(plugins.livereload());
}
PS: this is not a working example, and I dont know if you are using embedlr, but the point, is that you can watch, and use a callback to call another function with the files that are changing, and the manipulate only those files. Also, I supposed that your goal is to refresh the templates for your browser, but you manipulate as you like, save them on dest or do whatever you want.
Key point here is to show how to manipulate file that changes: callback of watch + custom function.
var jadeTask = function(path) {
path = path || loc.jade + '/*.jade';
if (/source/.test(path)) {
path = loc.jade + '/**/*.jade';
}
return gulp.src(path)
.pipe(changed(loc.markup, {extension: '.html'}))
.pipe(jade({
locals : json_array,
pretty : true
}))
.pipe(gulp.dest(loc.markup))
.pipe(connect.reload());
}
First install required plugins
gulp
express
gulp-jade
connect-livereload
tiny-lr
connect
then write the code
var gulp = require('gulp');
var express = require('express');
var path = require('path');
var connect = require("connect");
var jade = require('gulp-jade');
var app = express();
gulp.task('express', function() {
app.use(require('connect-livereload')({port: 8002}));
app.use(express.static(path.join(__dirname, '/dist')));
app.listen(8000);
});
var tinylr;
gulp.task('livereload', function() {
tinylr = require('tiny-lr')();
tinylr.listen(8002);
});
function notifyLiveReload(event) {
var fileName = require('path').relative(__dirname, event.path);
tinylr.changed({
body: {
files: [fileName]
}
});
}
gulp.task('jade', function(){
gulp.src('src/*.jade')
.pipe(jade())
.pipe(gulp.dest('dist'))
});
gulp.task('watch', function() {
gulp.watch('dist/*.html', notifyLiveReload);
gulp.watch('src/*.jade', ['jade']);
});
gulp.task('default', ['livereload', 'express', 'watch', 'jade'], function() {
});
find the example here at GitHub

How can I use factor-bundle with browserify programmatically?

I want to use factor-bundle to find common dependencies for my browserify entry points and save them out into a single common bundle:
https://www.npmjs.org/package/factor-bundle
The factor-bundle documentation makes it seem very easy to do on the command line, but I want to do it programmatically and I'm struggling to get my head around it.
My current script is this (I'm using reactify to transform react's jsx files too):
var browserify = require('browserify');
var factor = require('factor-bundle')
var glob = require('glob');
glob('static/js/'/**/*.{js,jsx}', function (err, files) {
var bundle = browserify({
debug: true
});
files.forEach(function(f) {
bundle.add('./' + f);
});
bundle.transform(require('reactify'));
// factor-bundle code goes here?
var dest = fs.createWriteStream('./static/js/build/common.js');
var stream = bundle.bundle().pipe(dest);
});
I'm trying to figure out how to use factor-bundle as a plugin, and specify the desired output file for each of the input files (ie each entry in files)
This answer is pretty late, so it's likely you've either already found a solution or a work around for this question. I'm answering this as it's quite similar to my question.
I was able to get this working by using factor-bundle as a browserify plugin. I haven't tested your specific code, but the pattern should be the same:
var fs = require('fs'),
browserify = require('browserify'),
factor = require('factor-bundle');
var bundle = browserify({
entries: ['x.js', 'y.js', 'z.js'],
debug: true
});
// Group common dependencies
// -o outputs the entry files without the common dependencies
bundle.plugin('factor-bundle', {
o: ['./static/js/build/x.js',
'./static/js/build/y.js',
'./static/js/build/z.js']
});
// Create Write Stream
var dest = fs.createWriteStream('./static/js/build/common.js');
// Bundle
var stream = bundle.bundle().pipe(dest);
The factor-bundle plugin takes output options o which need to have the same indexes as the entry files.
Unfortunately, I haven't figured out how to do anything else with these files after this point because I can't seem to access factor-bundle's stream event. So for minification etc, it might need to be done also via a browserify plugin.
I have created grunt-reactify to allow you to have a bundle file for a JSX file, in order to make it easier to work with modular React components.
All what you have to do is to specify a parent destination folder and the source files:
grunt.initConfig({
reactify: {
'tmp': 'test/**/*.jsx'
},
})

Best practice for minifying TypeScript modules

I'm using requirejs and AMD modules for my TypeScript project, with something like 20 different source files at the moment and likely to grow substantially. All of this works, but it's very slow to load all 20 files, so it would be better to have them minified. But because of how requirejs wants to load everything, it seems like it's going to require that I keep the modules in separate files - I don't think I can just take the generated module1.js and module2.js files and minify them into one file and then have requirejs load those without changing some code. (I could be wrong on this.)
The other way that I see to do this is to use the r.js file that requirejs provides to merge all the different files together in a way that still keeps requirejs happy. But r.js requires node.js, and I'd rather not introduce that as a dependency in my build process if there's any other way to do it.
So before I dive into this and try half a dozen different solutions - how are other folks approaching this with big projects?
What you could do is to implement a thin RequireJS shim to use in a minified build. Depending on how much of the RequireJS API you want to use, you could get by with very little. For simplicity you could also use named modules.
Say, while developing you use RequireJS to load your modules. When you want to make a minified build, you could simply include a simple loader in the minified file.
If you have files app.js, foo.js and bar.js as follows:
//from app.js
define("app", ["foo", "bar"], function(foo, bar) {
return {
run: function() { alert(foo + bar); }
}
});
//from foo.js
define("foo", [], function() {
return "Hello ";
});
//from bar.js
define("bar", [], function() {
return "World!";
});
And let's say you minify all those files together. At the top of the file you include the following shim:
//from your-require-shim.js
(function(exports) {
var modules = {};
var define = function(name, dependencies, func) {
modules[name] = {
name:name,
dependencies:dependencies,
func:func,
result:undefined
};
};
var require = function(name) {
var module = modules[name];
//if we have cached result -> return
if(module.result) { return module.result; }
var deps = [];
//resolve all dependencies
for(var i=0,len=module.dependencies.length;i<len;i++) {
var depName = module.dependencies[i];
var dep = modules[depName];
if(!dep.result) {
//resolve dependency
require(depName);
}
deps.push(dep.result);
}
module.result = module.func.apply(this, deps );
return module.result;
};
exports.require = require;
exports.define = define;
}(window));
And execute the module defined in app.js
require("app").run();
Like in this fiddle.
It's a crude PoC of course, but I'm sure you get the meaning.
If you are using ASP.NET MVC 4, you can make a bundle which will minify everything when you deploy to production in a set of files or in a folder. You'll find more info on bundles here.

Can I load multiple files with one require statement?

maybe this question is a little silly, but is it possible to load multiple .js files with one require statement? like this:
var mylib = require('./lib/mylibfiles');
and use:
mylib.foo(); //return "hello from one"
mylib.bar(): //return "hello from two"
And in the folder mylibfiles will have two files:
One.js
exports.foo= function(){return "hello from one";}
Two.js
exports.bar= function(){return "hello from two";}
I was thinking to put a package.json in the folder that say to load all the files, but I don't know how. Other aproach that I was thinking is to have a index.js that exports everything again but I will be duplicating work.
Thanks!!
P.D: I'm working with nodejs v0.611 on a windows 7 machine
First of all using require does not duplicate anything. It loads the module and it caches it, so calling require again will get it from memory (thus you can modify module at fly without interacting with its source code - this is sometimes desirable, for example when you want to store db connection inside module).
Also package.json does not load anything and does not interact with your app at all. It is only used for npm.
Now you cannot require multiple modules at once. For example what will happen if both One.js and Two.js have defined function with the same name?? There are more problems.
But what you can do, is to write additional file, say modules.js with the following content
module.exports = {
one : require('./one.js'),
two : require('./two.js'),
/* some other modules you want */
}
and then you can simply use
var modules = require('./modules.js');
modules.one.foo();
modules.two.bar();
I have a snippet of code that requires more than one module, but it doesn't clump them together as your post suggests. However, that can be overcome with a trick that I found.
function requireMany () {
return Array.prototype.slice.call(arguments).map(function (value) {
try {
return require(value)
}
catch (event) {
return console.log(event)
}
})
}
And you use it as such
requireMany("fs", "socket.io", "path")
Which will return
[ fs {}, socketio {}, path {} ]
If a module is not found, an error will be sent to the console. It won't break the programme. The error will be shown in the array as undefined. The array will not be shorter because one of the modules failed to load.
Then you can bind those each of those array elements to a variable name, like so:
var [fs, socketio, path] = requireMany("fs", "socket.io", "path")
It essentially works like an object, but assigns the keys and their values to the global namespace. So, in your case, you could do:
var [foo, bar] = requireMany("./foo.js", "./bar.js")
foo() //return "hello from one"
bar() //return "hello from two"
And if you do want it to break the programme on error, just use this modified version, which is smaller
function requireMany () {
return Array.prototype.slice.call(arguments).map(require)
}
Yes, you may require a folder as a module, according to the node docs. Let's say you want to require() a folder called ./mypack/.
Inside ./mypack/, create a package.json file with the name of the folder and a main javascript file with the same name, inside a ./lib/ directory.
{
"name" : "mypack",
"main" : "./lib/mypack.js"
}
Now you can use require('./mypack') and node will load ./mypack/lib/mypack.js.
However if you do not include this package.json file, it may still work. Without the file, node will attempt to load ./mypack/index.js, or if that's not there, ./mypack/index.node.
My understanding is that this could be beneficial if you have split your program into many javascript files but do not want to concatenate them for deployment.
You can use destructuring assignment to map an array of exported modules from require statements in one line:
const requires = (...modules) => modules.map(module => require(module));
const [fs, path] = requires('fs', 'path');
I was doing something similar to what #freakish suggests in his answer with a project where I've a list of test scripts that are pulled into a Puppeteer + Jest testing setup. My test files follow the naming convention testname1.js - testnameN.js and I was able use a generator function to require N number of files from the particular directory with the approach below:
const fs = require('fs');
const path = require('path');
module.exports = class FilesInDirectory {
constructor(directory) {
this.fid = fs.readdirSync(path.resolve(directory));
this.requiredFiles = (this.fid.map((fileId) => {
let resolvedPath = path.resolve(directory, fileId);
return require(resolvedPath);
})).filter(file => !!file);
}
printRetrievedFiles() {
console.log(this.requiredFiles);
}
nextFileGenerator() {
const parent = this;
const fidLength = parent.requiredFiles.length;
function* iterate(index) {
while (index < fidLength) {
yield parent.requiredFiles[index++];
}
}
return iterate(0);
}
}
Then use like so:
//Use in test
const FilesInDirectory = require('./utilities/getfilesindirectory');
const StepsCollection = new FilesInDirectory('./test-steps');
const StepsGenerator = StepsCollection.nextFileGenerator();
//Assuming we're in an async function
await StepsGenerator.next().value.FUNCTION_REQUIRED_FROM_FILE(someArg);

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