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'
},
})
Related
I've been trying to figure this out by myself, but had no success yet. I don't even know how to start researching for this (though I've tried some Google searchs already, to no avail), so I decided to ask this question here.
Is it possible to return multiple Vinyl files from a Through2 Object Stream?
My use case is this: I receive an HTML file via stream. I want to isolate two different sections of the files (using jQuery) and return them in two separate HTML files. I can do it with a single section (and a single resulting HTML file), but I have absolutely no idea on how I would do generate two different files.
Can anyone give me a hand here?
Thanks in advance.
The basic approach is something like this:
Create as many output files from your input file as you need using the clone() function.
Modify the .contents property of each file depending on what you want to do. Don't forget that this is a Buffer, not a String.
Modify the .path property of each file so your files don't overwrite each other. This is an absolute path so use something like path.parse() and path.join() to make things easier.
Call this.push() from within the through2 transform function for every file you have created.
Here's a quick example that splits a file test.txt into two equally large files test1.txt and test2.txt:
var gulp = require('gulp');
var through = require('through2').obj;
var path = require('path');
gulp.task('default', function () {
return gulp.src('test.txt')
.pipe(through(function(file, enc, cb) {
var c = file.contents.toString();
var f = path.parse(file.path);
var file1 = file.clone();
var file2 = file.clone();
file1.contents = new Buffer(c.substring(0, c.length / 2));
file2.contents = new Buffer(c.substring(c.length / 2));
file1.path = path.join(f.dir, f.name + '1' + f.ext);
file2.path = path.join(f.dir, f.name + '2' + f.ext);
this.push(file1);
this.push(file2);
cb();
}))
.pipe(gulp.dest('out'));
});
I've successfully got Browserify to compile my JavaScript entry files, but I want to utilise the Remapify plugin so as to not have to specify the full relative path upon requiring a module every time.
For example:
require('components/tabs.js')
Rather than:
require('../../components/tabs/tabs.js').
But I cannot get the shorter module references to map to the corresponding file... "Error: Cannot find module [specified_ref] from [file]".
Have I misconfigured Remapify, or is there something wrong with my wider Browserify setup? I am new to Broswerify and Gulp having previously used Require.js and Grunt. Any help would be greatly appreciated. Please let me know if you need any more information about my setup.
If alternatively you can recommend an alternative Gulp task file that will do all of this, thereby throwing my current task out the window, by all means. I wasn't able to find many Browserify + Remapify examples.
Directory Structure
I have my modules (components) in the following directory: './src/components', so for example: './src/components/tabs/tabs.js'.
I am requiring these modules in a JS file for a given page of the app, which are in: './src/pages', so for example, './src/pages/portfolio/portfolio.js'.
Gulp Browserify Task
var gulp = require('gulp');
var config = require('../config');
var browserify = require('browserify');
var remapify = require('remapify');
var source = require('vinyl-source-stream');
var glob = require('glob');
var browserSync = require('browser-sync');
gulp.task('browserify', function(){
var entries = glob.sync(config.src.pages + '/**/*.js');
return browserify({
entries: entries,
debug: true
})
// (Remapify:)
.plugin(remapify, [{ src: config.src.components + '/**/*.js', expose: 'components', cwd: config.srcDir }])
.bundle()
.pipe(source('app.js'))
.pipe(gulp.dest(config.build.js))
.pipe(browserSync.reload({ stream: true }));
});
Page.js
'use strict';
var tabs = require('components/tabs.js'); // (Doesn't work, but I want it to)
// var tabs = require('../../components/tabs/tabs.js'); // (Does work)
Remapify has all sorts of problems. I suggest giving my pathmodify plugin a shot.
For your situation usage would look something like:
var pathmod = require('pathmodify');
// ...
.plugin(pathmod(), {mods: [
pathmod.mod.dir('components', '/path/to/src/components'),
]})
I want to use an yeoman generator inside a NodeJS project
I installed yeoman-generatorand generator-git (the generator that I want use) as locally dependency, and, at this moment my code is like this:
var env = require('yeoman-generator')();
var path = require('path');
var gitGenerator = require('generator-git');
var workingDirectory = path.join(process.cwd(), 'install_here/');
generator = env.create(gitGenerator);
obviously the last line doesn't work and doesn't generate the scaffold.
The question: How to?
Importantly, I want to stay in local dependency level!
#simon-boudrias's solution does work, but after I changed the process.chdir(), this.templatePath() and this.destinationPath() returns same path.
I could have use this.sourcePath() to tweak the template path, but having to change this to each yeoman generator is not so useful. After digging to yo-cli, I found the following works without affecting the path.
var env = require('yeoman-environment').createEnv();
env.lookup(function() {
env.run('generator-name');
});
env.create() only instantiate a generator - it doesn't run it.
To run it, you could call generator.run(). But that's not ideal.
The best way IMO would be this way:
var path = require('path');
var env = require('yeoman-generator')();
var gitGenerator = require('generator-git');
// Optionnal: look every generator in your system. That'll allow composition if needed:
// env.lookup();
env.registerStub(gitGenerator, 'git:app');
env.run('git:app');
If necessary, make sure to process.chdir() in the right directory before launching your generator.
Relevant documentation on the Yeoman Environment class can be found here: http://yeoman.io/environment/Environment.html
Also see: http://yeoman.io/authoring/integrating-yeoman.html
The yeoman-test module is also very useful if you want to pass predefined answers to your prompts. This worked for me.
var yeomanTest = require('yeoman-test');
var answers = require('from/some/file.json');
var context = yeomanTest.run(path.resolve('path/to/generator'));
context.settings.tmpdir = false; // don't run in tempdir
context.withGenerators([
'paths/to/subgenerators',
'more/of/them'
])
.withOptions({ // execute with options
'skip-install': true,
'skip-sdk': true
})
.withPrompts(answers) // answer prompts
.on('end', function () {
// do some stuff here
});
I have a bunch of html files in a partials directory. Using gulp js, I want to minify and rename these files to .min.html. Please show me how to achieve this.
See here, using gulp-rename, if you just want to rename the files.
Something in the line of below should do:
var rename = require('gulp-rename');
gulp.src("./partials/**/*.hmtl")
.pipe(rename(function (path) {
path.suffix += ".min";
}))
.pipe(gulp.dest("./dist"));
In order to minify, you can use gulp-htmlmin. Pretty straightforward from the documentation:
var htmlmin = require('gulp-htmlmin');
gulp.task('compress', function() {
gulp.src('./partials/**/*.html')
.pipe(htmlmin())
.pipe(gulp.dest('dist'))
});
You can certainly combine the two to obtain the desired effect.
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.