Say, for example, you are building a project on Backbone or whatever and you need to load scripts in a certain order, e.g. underscore.js needs to be loaded before backbone.js.
How do I get it to concat the scripts so that they’re in order?
// JS concat, strip debugging and minify
gulp.task('scripts', function() {
gulp.src(['./source/js/*.js', './source/js/**/*.js'])
.pipe(concat('script.js'))
.pipe(stripDebug())
.pipe(uglify())
.pipe(gulp.dest('./build/js/'));
});
I have the right order of scripts in my source/index.html, but since files are organized by alphabetic order, gulp will concat underscore.js after backbone.js, and the order of the scripts in my source/index.html does not matter, it looks at the files in the directory.
So does anyone have an idea on this?
Best idea I have is to rename the vendor scripts with 1, 2, 3 to give them the proper order, but I am not sure if I like this.
As I learned more I found Browserify is a great solution, it can be a pain at first but it’s great.
I had a similar problem recently with Grunt when building my AngularJS app. Here's a question I posted.
What I ended up doing is to explicitly list the files in order in the grunt config. The config file will then look like this:
[
'/path/to/app.js',
'/path/to/mymodule/mymodule.js',
'/path/to/mymodule/mymodule/*.js'
]
Grunt is able to figure out which files are duplicates and not include them. The same technique will work with Gulp as well.
Another thing that helps if you need some files to come after a blob of files, is to exclude specific files from your glob, like so:
[
'/src/**/!(foobar)*.js', // all files that end in .js EXCEPT foobar*.js
'/src/js/foobar.js',
]
You can combine this with specifying files that need to come first as explained in Chad Johnson's answer.
I have used the gulp-order plugin but it is not always successful as you can see by my stack overflow post gulp-order node module with merged streams. When browsing through the Gulp docs I came across the streamque module which has worked quite well for specifying order of in my case concatenation. https://github.com/gulpjs/gulp/blob/master/docs/recipes/using-multiple-sources-in-one-task.md
Example of how I used it is below
var gulp = require('gulp');
var concat = require('gulp-concat');
var handleErrors = require('../util/handleErrors');
var streamqueue = require('streamqueue');
gulp.task('scripts', function() {
return streamqueue({ objectMode: true },
gulp.src('./public/angular/config/*.js'),
gulp.src('./public/angular/services/**/*.js'),
gulp.src('./public/angular/modules/**/*.js'),
gulp.src('./public/angular/primitives/**/*.js'),
gulp.src('./public/js/**/*.js')
)
.pipe(concat('app.js'))
.pipe(gulp.dest('./public/build/js'))
.on('error', handleErrors);
});
With gulp-useref you can concatenate every script declared in your index file, in the order in which you declare it.
https://www.npmjs.com/package/gulp-useref
var $ = require('gulp-load-plugins')();
gulp.task('jsbuild', function () {
var assets = $.useref.assets({searchPath: '{.tmp,app}'});
return gulp.src('app/**/*.html')
.pipe(assets)
.pipe($.if('*.js', $.uglify({preserveComments: 'some'})))
.pipe(gulp.dest('dist'))
.pipe($.size({title: 'html'}));
});
And in the HTML you have to declare the name of the build file you want to generate, like this:
<!-- build:js js/main.min.js -->
<script src="js/vendor/vendor.js"></script>
<script src="js/modules/test.js"></script>
<script src="js/main.js"></script>
In your build directory you will have the reference to main.min.js which will contain vendor.js, test.js, and main.js
The sort-stream may also be used to ensure specific order of files with gulp.src. Sample code that puts the backbone.js always as the last file to process:
var gulp = require('gulp');
var sort = require('sort-stream');
gulp.task('scripts', function() {
gulp.src(['./source/js/*.js', './source/js/**/*.js'])
.pipe(sort(function(a, b){
aScore = a.path.match(/backbone.js$/) ? 1 : 0;
bScore = b.path.match(/backbone.js$/) ? 1 : 0;
return aScore - bScore;
}))
.pipe(concat('script.js'))
.pipe(stripDebug())
.pipe(uglify())
.pipe(gulp.dest('./build/js/'));
});
I just add numbers to the beginning of file name:
0_normalize.scss
1_tikitaka.scss
main.scss
It works in gulp without any problems.
I have my scripts organized in different folders for each package I pull in from bower, plus my own script for my app. Since you are going to list the order of these scripts somewhere, why not just list them in your gulp file? For new developers on your project, it's nice that all your script end-points are listed here. You can do this with gulp-add-src:
gulpfile.js
var gulp = require('gulp'),
less = require('gulp-less'),
minifyCSS = require('gulp-minify-css'),
uglify = require('gulp-uglify'),
concat = require('gulp-concat'),
addsrc = require('gulp-add-src'),
sourcemaps = require('gulp-sourcemaps');
// CSS & Less
gulp.task('css', function(){
gulp.src('less/all.less')
.pipe(sourcemaps.init())
.pipe(less())
.pipe(minifyCSS())
.pipe(sourcemaps.write('source-maps'))
.pipe(gulp.dest('public/css'));
});
// JS
gulp.task('js', function() {
gulp.src('resources/assets/bower/jquery/dist/jquery.js')
.pipe(addsrc.append('resources/assets/bower/bootstrap/dist/js/bootstrap.js'))
.pipe(addsrc.append('resources/assets/bower/blahblah/dist/js/blah.js'))
.pipe(addsrc.append('resources/assets/js/my-script.js'))
.pipe(sourcemaps.init())
.pipe(concat('all.js'))
.pipe(uglify())
.pipe(sourcemaps.write('source-maps'))
.pipe(gulp.dest('public/js'));
});
gulp.task('default',['css','js']);
Note: jQuery and Bootstrap added for demonstration purposes of order. Probably better to use CDNs for those since they are so widely used and browsers could have them cached from other sites already.
Try stream-series. It works like merge-stream/event-stream.merge() except that instead of interleaving, it appends to the end. It doesn't require you to specify the object mode like streamqueue, so your code comes out cleaner.
var series = require('stream-series');
gulp.task('minifyInOrder', function() {
return series(gulp.src('vendor/*'),gulp.src('extra'),gulp.src('house/*'))
.pipe(concat('a.js'))
.pipe(uglify())
.pipe(gulp.dest('dest'))
});
merge2 looks like the only working and maintained ordered stream merging tool at the moment.
Update 2020
The APIs are always changing, some libraries become unusable or contain vulnerabilities, or their dependencies contain vulnerabilities, that are not fixed for years. For text files manipulations you'd better use custom NodeJS scripts and popular libraries like globby and fs-extra along with other libraries without Gulp, Grunt, etc wrappers.
import globby from 'globby';
import fs from 'fs-extra';
async function bundleScripts() {
const rootPaths = await globby('./source/js/*.js');
const otherPaths = (await globby('./source/**/*.js'))
.filter(f => !rootFiles.includes(f));
const paths = rootPaths.concat(otherPaths);
const files = Promise.all(
paths.map(
// Returns a Promise
path => fs.readFile(path, {encoding: 'utf8'})
)
);
let bundle = files.join('\n');
bundle = uglify(bundle);
bundle = whatever(bundle);
bundle = bundle.replace(/\/\*.*?\*\//g, '');
await fs.outputFile('./build/js/script.js', bundle, {encoding: 'utf8'});
}
bundleScripts.then(() => console.log('done');
An alternative method is to use a Gulp plugin created specifically for this problem. https://www.npmjs.com/package/gulp-ng-module-sort
It allows you to sort your scripts by adding in a .pipe(ngModuleSort()) as such:
var ngModuleSort = require('gulp-ng-module-sort');
var concat = require('gulp-concat');
gulp.task('angular-scripts', function() {
return gulp.src('./src/app/**/*.js')
.pipe(ngModuleSort())
.pipe(concat('angularAppScripts.js))
.pipe(gulp.dest('./dist/));
});
Assuming a directory convention of:
|——— src/
| |——— app/
| |——— module1/
| |——— sub-module1/
| |——— sub-module1.js
| |——— module1.js
| |——— module2/
| |——— sub-module2/
| |——— sub-module2.js
| |——— sub-module3/
| |——— sub-module3.js
| |——— module2.js
| |——— app.js
Hope this helps!
For me I had natualSort() and angularFileSort() in pipe which was reordering the files. I removed it and now it works fine for me
$.inject( // app/**/*.js files
gulp.src(paths.jsFiles)
.pipe($.plumber()), // use plumber so watch can start despite js errors
//.pipe($.naturalSort())
//.pipe($.angularFilesort()),
{relative: true}))
I just use gulp-angular-filesort
function concatOrder() {
return gulp.src('./build/src/app/**/*.js')
.pipe(sort())
.pipe(plug.concat('concat.js'))
.pipe(gulp.dest('./output/'));
}
I'm in a module environnement where all are core-dependents using gulp.
So, the core module needs to be appended before the others.
What I did:
Move all the scripts to an src folder
Just gulp-rename your core directory to _core
gulp is keeping the order of your gulp.src, my concat src looks like this:
concat: ['./client/src/js/*.js', './client/src/js/**/*.js', './client/src/js/**/**/*.js']
It'll obviously take the _ as the first directory from the list (natural sort?).
Note (angularjs):
I then use gulp-angular-extender to dynamically add the modules to the core module.
Compiled it looks like this:
angular.module('Core', ["ui.router","mm.foundation",(...),"Admin","Products"])
Where Admin and Products are two modules.
if you would like to order third party libraries dependencies, try wiredep. This package basically checks each package dependency in bower.json then wire them up for you.
I tried several solutions from this page, but none worked. I had a series of numbered files which I simply wanted be ordered by alphabetical foldername so when piped to concat() they'd be in the same order. That is, preserve the order of the globbing input. Easy, right?
Here's my specific proof-of-concept code (print is just to see the order printed to the cli):
var order = require('gulp-order');
var gulp = require('gulp');
var print = require('gulp-print').default;
var options = {};
options.rootPath = {
inputDir: process.env.INIT_CWD + '/Draft',
inputGlob: '/**/*.md',
};
gulp.task('default', function(){
gulp.src(options.rootPath.inputDir + options.rootPath.inputGlob, {base: '.'})
.pipe(order([options.rootPath.inputDir + options.rootPath.inputGlob]))
.pipe(print());
});
The reason for the madness of gulp.src? I determined that gulp.src was running async when I was able to use a sleep() function (using a .map with sleeptime incremented by index) to order the stream output properly.
The upshot of the async of src mean dirs with more files in it came after dirs with fewer files, because they took longer to process.
In my gulp setup, I'm specifying the vendor files first and then specifying the (more general) everything, second. And it successfully puts the vendor js before the other custom stuff.
gulp.src([
// vendor folder first
path.join(folder, '/vendor/**/*.js'),
// custom js after vendor
path.join(folder, '/**/*.js')
])
Apparently you can pass in the "nosort" option to gulp.src gulp.src.
Related
So, I'm moving from grunt to gulp (or trying to anyway), and I'm having trouble getting gulp to do what I'm doing in grunt. Specifically the $templateCache stuff.
My angular app is broken up into several components/modules. Each module contains everything it needs to run (controllers, directives, partials, scss, etc.).
Using Grunt, I've been able to boil each module down into 5 files:
module.min.css // all module scss files compiled and concatenated
module.min.js // all module controllers, directives, services, etc. concatenated
module.tpls.min.js // all partials in $templateCache for this module
module.mocks.min.js // all unit test mock objects for this module
module.specs.min.js // all unit test specs for this module
This has worked really well for 2 years now and been a cornerstone of my modular architecture. My only reasons to try out gulp was 1) Curiosity, 2) My grunt file is getting kinda hairy as we add in deployment and environment specific stuff and so far gulp has really slimmed that down.
For the most part, I've figured out how to do all my grunt tasks in gulp, but I'm having trouble figuring out how to generate a template cache file for each module. All the gulp-ng|angular-templates|templatecache plugins take all my partials and create one file. I'd like to take all my files under module/partials/*.html and create a single module.tpls.min.js; and do that for each module.
This was actually a problem with grunt too, but I figured it out with grunt.file.expand().forEach() like this:
grunt.registerTask('prepModules', '...', function(){
// loop through our modules directory and create subtasks
// for each module, modifying tasks that affect modules.
grunt.file.expand("src/js/modules/*").forEach(function (dir) {
// get the module name by looking at the directory we're in
var mName = dir.substr(dir.lastIndexOf('/') + 1);
// add ngtemplate subtasks for each module, turning
// all module partials into $templateCache objects
ngtemplates[mName] = {
module: mName,
src: dir + "/partials/**/*.html",
dest: 'dev/modules/' + mName + '/' + mName + '.tpls.min.js'
};
grunt.config.set('ngtemplates', ngtemplates);
});
});
My current gulp for this same task:
var compileTemplates = gulp.src('./src/js/modules/**/partials/*.html', {base:'.'})
.pipe(ngTemplates())
.pipe(gulp.dest('.'));
I've only really looked at the options, but none of them seemed to do what I wanted. They were all around changing the file name, or the final destination of the file, or a module name, or whatever else; nothing that said anything about doing it for only the directory it happens to be in.
I had thought about using gulp-rename because it worked well for me when doing the CSS compilation:
var compileScss = gulp.src('./src/js/modules/**/scss/*.scss', {base:'.'})
.pipe(sass({includePaths: ['./src/scss']}))
.pipe(rename(function(path){
path.dirname = path.dirname.replace(/scss/,'css');
}))
.pipe(gulp.dest('.'));
However, when I pipe rename() after doing ngTemplates() it only has the path of the final output file (one log entry). When you console.log() path after sass(), it has all the paths of all the files that it found (lots of log entries).
Any ideas? Thanks!
This SO post has the correct answer, but the wasn't coming up in my searches for this specific usage. I was going to vote to close my question, but since someone else might search using my own specific terms (since I did), it seems more appropriate to leave it alone and just redirect to the original question as well as show how I solved my own particular problem.
var fs = require('fs');
var ngTemplates = require('gulp-ng-templates');
var rename = require('gulp-rename');
var modulesDir = './src/js/modules/';
var getModules = function(dir){
return fs.readdirSync(dir)
.filter(function(file){
return fs.statSync(path.join(dir, file)).isDirectory();
});
};
gulp.task('default', function(){
var modules = getModules(modulesDir);
var moduleTasks = modules.map(function(folder){
// get all partials for this module
// parse into $templateCache file
// rename to be /dev/modules/_____/______.tpls.min.js
return gulp.src(modulesDir + folder + '/partials/*.html', {basedir:'.'})
.pipe(ngTemplates({module:folder}))
.pipe(rename(function(path){
path.dirname = './dev/apps/' + folder + '/';
path.basename = folder + '.tpls.min';
}))
.pipe(gulp.dest('.'));
});
});
It's essentially like the tasks per folder recipe but with a change to use gulp-ng-templates. I'll probably be using this same pattern for my SCSS and JS now that I'm more aware of it.
Seems like the gulp equivalent of grunt.file.expand().forEach().
Whenever I deal with scss/sass for gulp tasks, I will only put one scss file as the source parameter. This parameter file is composed of a list of imports. This way you don't need to rely on gulp to concat the scss file contents for you.
//in gulpfile
gulp.src('./src/js/modules/**/scss/main.scss', {base:'.'})
//in main.scss
#import 'a', 'b', 'c';
a, b, and c would represent your other scss files.
I have a gulp starter kit for my project, however, I want to use gulp-load-plugins to for devDependencies of package.json file. My file structure is
ProjectName
Gulp
-Tasks
-broswerify.js
-browserSync.js
-jade.js
-lint.js
Gulpfile.js
config.json
package.json
Gulpfile.js
var requireDir = require('require-dir');
var dir = requireDir('./gulp/tasks', {recurse: true});
jade.js (Which is working as expected using gulp-load-plugins)
var gulp = require('gulp');
var config = require('../../config.json');
var plugins = require('gulp-load-plugins')();
gulp.task('jade', function(){
return gulp.src(config.jade.src)
.pipe(plugins.jade())
.pipe(gulp.dest(config.jade.build))
});
browsersync.js (which is not working using gulp-load-plugins)
var gulp = require('gulp');
var config = require('../../config.json').browsersync;
var plugins = require('browsersync'); // works
//var plugins = require('gulp-load-plugins')(); // it doesn't works.
gulp.task('browsersync', function () {
plugins.browserSync.init(config); // (browsersync required)
//plugins.browserSync.init(config) it doesn't work. (gulp-load-plugins required)
});
I would like to know that if there is a better way to do that?
Why would you wan't to use gulp-load-plugins if you have a seperate file for each plugin?
This is how i load gulp-load-plugins :
$ = require('gulp-load-plugins')({
pattern: ['gulp-*', 'gulp.*'],
replaceString: /\bgulp[\-.]/,
lazy: true,
camelize: true
}),
Here is an example of a revision plugin:
// revision styles
gulp.task('rev-styles', function () {
return gulp.src(dev.css)
.pipe($.rev())
.pipe($.cssmin())
.pipe(gulp.dest(dist.css))
.pipe($.filesize())
.pipe($.rev.manifest({merge: true}))
.pipe(gulp.dest('./'))
//rev replace
.on('end', function() {
return gulp.src(['./rev-manifest.json', 'dist/*.html'])
.pipe($.revCollector({
replaceReved: true,
dirReplacements: {
'css': 'css'
}
}))
.pipe(gulp.dest(dist.dist))
});
});
As you can see all my pipes are called .pipe($.pluginName()) meaning $ stands for gulp- . If you have a plugin named gulp-name-secondname you call it like this: .pipe($.nameSecondname()) . Top were i require gulp-load-plugins i have camelize set to true . Lazy loading loads only the plugins you use not all of them .
As a side note i strongly recommend not separating plugins in diffrent files but you can modulate them, meaning separating important tasks in separate files like compilation file , optimization file , build file, etc .
This might help you understand gulp file separation better http://macr.ae/article/splitting-gulpfile-multiple-files.html
Careful with gulp-load-plugins because it slows your tasks , for example i run gulp-webserver , when i use it with gulp-load-plugins the task finishes after 200ms versus 20ms if i use it normally. So don't use with everything, play with it see how much performance you lose on each task and prioritize.
I have used gulp-load-plugins but found that it mainly adds complexity and obscures my code. At also makes it harder to understand for people less familiar with Gulp. It looks cleaner and easier to understand to have all modules explicitly declared at the top.
I am trying to merge css and scss files into a main.css file that goes in my build directory.
Its working, but not in the right order. The style attributes from the scss files need to be in the bottom of the main.css file so they overrule the rest.
my Gulp task looks like this:
//CSS
gulp.task('css', function () {
var cssTomincss = gulp.src(['dev/css/reset.css', 'dev/css/style.css','dev/css/typography.css', 'dev/css/sizes.css']);
var cssFromscss = gulp.src(['dev/css/*.scss'])
.pipe(sass());
return es.merge(cssTomincss, cssFromscss)
.pipe(concat('main.css'))
.pipe(minifyCSS())
.pipe(gulp.dest('build/css'))
});
I am defining the sources first with variables. I am using the gulp-sass plugin to convert the scss file into normal css (.pipe(sass)) and later merging the two with the es.merge function and concatenating them into main.css.
The problem is that the style attributes van the .scss files end up somewhere in the top end of the main.css file. I need them to be at the bottom. So they need to be concatenated at the bottom.
Any clue on how to do this?
Try streamqueue.
var streamqueue = require('streamqueue');
gulp.task('css', function () {
return streamqueue({ objectMode: true },
gulp.src(['dev/css/reset.css', 'dev/css/style.css', 'dev/css/typography.css', 'dev/css/sizes.css']),
gulp.src(['dev/css/*.scss']).pipe(sass())
)
.pipe(concat('main.css'))
.pipe(minifyCSS())
.pipe(gulp.dest('build/css'))
});
This cheatsheet will help you. PDF is here.
It seems that the plugin gulp-order fits perfectly well in your case.
It allows you to re-order the passed stream with your own glob pattern, for example based on your code :
return es.merge(cssTomincss, cssFromscss)
.pipe(order([
'dev/css/reset.css',
'dev/css/style.css',
'dev/css/typography.css',
'dev/css/sizes.css',
'dev/css/*.css',
]))
.pipe(concat('main.css'))
.pipe(minifyCSS())
.pipe(gulp.dest('build/css'))
One drawback of this is that you have to re-declare your globs, but you can get around by assign your globs to a value and then concat them in you order pipe, much cleaner.
You may have to set the base option to . of gulp-order as stated in their Readme if the files were not ordered correctly.
One another way would be to use stream-series, basically the same as event-stream, but the order of your stream is preserved, and you don't have to rewrite your globs.
I tried gulp-order without success: the order somehow wasn't taken into account.
The solution which worked for me was using stream-series, mentioned by Aperçu.
return streamSeries(
cssTomincss,
cssFromscss)
.pipe(concat('main.css'))
.pipe(minifyCSS())
.pipe(gulp.dest('build/css'));
I failed with all provided answers, they produced some silent errors. Finally merge2 worked for me (seems like there was gulp-merge and later the project was renamed into merge2). I'm not sure why there is a need in streamify plugin, e.g. streams created with Rollup may produce "stream-not-supported-errors" with gulp-concat, gulp-uglify or gulp-insert.
const mergeStreams = require('merge2');
const streamify = require('streamify');
...
gulp.task('build', () => {
const streams = sources.map(createJSFile);
return mergeStreams(...streams)
.pipe(streamify(concat('bundle.js')))
.pipe(streamify(uglify()))
.pipe(gulp.dest('./dist'));
});
I'm experimenting with using gulpjs instead of grunt for a project. I'm attempting to use gulp filter to ignore vendor libraries when running jsHint on my code. I've based my code off of the code from the readme's example, but the files have not been filtered.
I'm running node 0.10.26, gulp 3.8.0,and gulp filter 0.4.1
I'm trying to run jshint on a directory wcui/app/js that contains many other directories of JS files, with about 120 js files total. I want to exclude the vendor directory only.
My code looks like this:
var gulp = require('gulp');
var gulpFilter = require('gulp-filter');
var jshint = require('gulp-jshint');
var srcs = {
scripts: ['wcui/app/js/**/*.js'],
styles: ['wcui/app/css/**/*.less','wcui/app/css/**/*.css']
};
var dests = {
scripts: 'wcui/static/js/',
styles: 'wcui/static/css/'
};
gulp.task('scripts', function() {
var filter = gulpFilter('!wcui/app/js/vendor');
return gulp.src(srcs.scripts)
.pipe(filter)
.pipe(jshint())
.pipe(jshint.reporter('jshint-stylish'))
.pipe(filter.restore)
.pipe(gulp.dest(dests.scripts));
});
gulp.task('styles', function() {
return gulp.src(srcs.styles)
.pipe(gulp.dest(dests.styles));
});
gulp.task('dev',['scripts','styles']);
Right now running gulp dev does the same thing it did before I added the filter, linting every js file. How can I change this to make it filter correctly? The gulp example had the src in the format 'wcui/app/js/*.js' but when I admit the ** glob, I don't get subdirectories at all. Other than that I think I'm following the readme to the letter (with changes for my particular task).
For readers that have a more up-to-date version of gulp-filter (release at the time of writing is 1.0.0)
The release of version 0.5.0 of gulp-filter introduced multimatch 0.3.0 which come with a breaking change.
Breaking change
Using a negate ! as the first pattern no longer matches anything.
Workaround: ['*', '!cake']
Basically, what it means is you need to replace
var filter = gulpFilter('!wcui/app/js/vendor');
with
var filter = gulpFilter(['*', '!wcui/app/js/vendor']);
and you are good to go.
Also, as noted in the comment by MildlySerious, you should have .pipe(filter.restore()) instead of .pipe(filter.restore)
Use filter like this gulpFilter(['*', '!app/vendor'])
I have browserify bundling up files and it's working great. But what if I need to generate multiple bundles?
I would like to end up with dist/appBundle.js and dist/publicBundle.js
gulp.task("js", function(){
return browserify([
"./js/app.js",
"./js/public.js"
])
.bundle()
.pipe(source("bundle.js"))
.pipe(gulp.dest("./dist"));
});
Obviously this isn't going to work since I am only specifying one output (bundle.js). I can accomplish this by repeating the above statement like so (but it doesn't feel right, because of the repetition):
gulp.task("js", function(){
browserify([
"./js/app.js"
])
.bundle()
.pipe(source("appBundle.js"))
.pipe(gulp.dest("./dist"));
browserify([
"./js/public.js"
])
.bundle()
.pipe(source("publicBundle.js"))
.pipe(gulp.dest("./dist"));
});
Is there a better way to tackle this? Thanks!
I don't have a good environment to test this in right now, but my guess is that it would look something like:
gulp.task("js", function(){
var destDir = "./dist";
return browserify([
"./js/app.js",
"./js/public.js"
])
.bundle()
.pipe(source("appBundle.js"))
.pipe(gulp.dest(destDir))
.pipe(rename("publicBundle.js"))
.pipe(gulp.dest(destDir));
});
EDIT: I just realized I mis-read the question, there should be two separate bundles coming from two separate .js files. In light of that, the best alternative I can think of looks like:
gulp.task("js", function(){
var destDir = "./dist";
var bundleThis = function(srcArray) {
_.each(srcArray, function(source) {
var bundle = browserify(["./js/" + source + ".js"]).bundle();
bundle.pipe(source(source + "Bundle.js"))
.pipe(gulp.dest(destDir));
});
};
bundleThis(["app", "public"]);
});
gulp.task("js", function (done) {
[
"app",
"public",
].forEach(function (entry, i, entries) {
// Count remaining bundling operations to track
// when to call done(). Could alternatively use
// merge-stream and return its output.
entries.remaining = entries.remaining || entries.length;
browserify('./js/' + entry + '.js')
.bundle()
// If you need to use gulp plugins after bundling then you can
// pipe to vinyl-source-stream then gulp.dest() here instead
.pipe(
require('fs').createWriteStream('./dist/' + entry + 'Bundle.js')
.on('finish', function () {
if (! --entries.remaining) done();
})
);
});
});
This is similar to #urban_racoons answer, but with some improvements:
That answer will fail as soon as you want the task to be a dependency of another task in gulp 3, or part of a series in gulp 4. This answer uses a callback to signal task completion.
The JS can be simpler and doesn't require underscore.
This answer is based on the premise of having a known list of entry files for each bundle, as opposed to, say, needing to glob a list of entry files.
Multiple bundles with shared dependencies
I recently added support for multiple bundles with shared dependencies to https://github.com/greypants/gulp-starter
Here's the array of browserify config objects I pass to my browserify task. At the end of that task, I iterate over each config, browserifying all the things.
config.bundleConfigs.forEach(browserifyThis);
browserifyThis takes a bundleConfig object, and runs browserify (with watchify if dev mode).
This is the bit that sorts out shared dependencies:
// Sort out shared dependencies.
// b.require exposes modules externally
if(bundleConfig.require) b.require(bundleConfig.require)
// b.external excludes modules from the bundle, and expects
// they'll be available externally
if(bundleConfig.external) b.external(bundleConfig.external)
This browserify task also properly reports when all bundles are finished (the above example isn't returning streams or firing the task's callback), and uses watchify when in devMode for super fast recompiles.
Brian FitzGerald's last comment is spot on. Remember that it's just JavaScript!