So I'm having a slight problem with producing production ready scripts for my project. I'm using gulp to concatenate and minify my css and js, and while the css is working fine the gulp js function isn't generating my final file. Please refer to my code below:
gulp.task('js', function() {
return gulp.src([source + 'js/app/**/*.js'])
.pipe(concat('development.js'))
.pipe(gulp.dest(source + 'js'))
.pipe(rename({
basename: 'production',
suffix: '-min',
}))
.pipe(uglify())
.pipe(gulp.dest(source + 'js/'))
.pipe(notify({ message: 'Scripts task complete', onLast: true }));
});
If anyone has encountered a similar problem or has any tips it would be much appreciated :)
There is nothing wrong with your gulpfile. I tested it and it works perfectly.
The only thing I can guess is that your source is not set correctly. Did you forget the trailing slash '/' ?
I would suggest 2 things to figure it out. Include node path library to check where source is actually pointing to like this:
var path = require('path');
// in gulp task ...
path.resolve(path.resolve(source + 'js/app'));
Make sure it points where you think it does.
Secondly, you could use gulp-debug to establish that any files are found:
npm install gulp-debug
Then
var debug = require('gulp-debug');
// in gulp task ...
return gulp.src([source + 'js/app/**/*.js'])
.pipe(concat('development.js'))
.pipe(debug())
.pipe(gulp.dest(source + 'js'))
.pipe(debug())
// etc.
Good luck!
Based on additional infomation in the comments I realise you are generating JS files in a separate process ...
gulp is asynchronous by default. What this boils down to is that all functions try to run at the same time - if you want a specific order it must be by design. This is great because it's very fast but can be a headache to work with.
Problem
Here is what's basically happening:
// SOME TASK THAT SHOULD BE RUN FIRST
gulp.task('copy-vendor-files-to-tempfolder', function (done) {
// copy files to vendor folder
done()
})
// SOME TASKS THAT DEPEND ON FIRST TASK
gulp.task('complile-styles', function () { /* independent task */ })
gulp.task('concat-vendor-files', function () { /* concat files in vendor folder. depends on vendor files existing */ })
// GENERAL TASK WHICH STARTS OTHERS
gulp.task('ready', ['copy-vendor-files-to-tempfolder', 'compile-styles', 'concat-vendor-files])
When you try to run:
$ gulp ready
GULP TASK WILL FAIL! Folder is being created at the same time!!
NOWHERE TO COPY FILES!
Solution
There are many solutions but the following module has come in handy for me again and again:
npm install run-sequence
Then in your gulpfile.js:
var runSequence = require('run-sequence')
gulp.task('ready', function (done) {
runSequence(
'create-folders', // do this first,
[
'copy-css-files',
'copy-html-files'
], // do these AFTER but in parallel
done // callback when ready
)
})
This will guarantee the folder exists when you try to run the other functions.
In your specific case, you should make sure the task that concatenates the JS files is run after the task that copies them out of vendor.
Note: I'm leaving other answer because it contains useful help for debugging similar issues.
HTH!
I've started using React with Flux architecture for full functional frontend application, and I really liked approach JSX and Flux , but the main issue is that when I'm building JSX files using Gulp, Babel and Uglyfy I'm getting about 1mb minified JS file, without minified mode it is giving almost 8mb of JS file.
And that's not the end !! for making AJAX requests React don't have built in functionality, so I need also include jQuery
I's working well, development process is faster and code is nicer than with other frameworks thanks to JSX. But how to make production files smaller ?
I'm including just a few libs Dispatcher and EventEmmiter for Flux architecture. So it's not about having unused libs in my code. I think it is because I'm writing JSX and now I have HTML+JS in one single file.
What is the best practice to split files or make JS output more smaller ?
Thanks !
There are some steps to reduce production size:
Use production version of ReactJS, which includes extra performance optimizations and strips all error messages.
You don't have to include the whole jQuery library to just use Ajax, I suggest use other lightweight library for handling ajax only e.g. reqwest or superagent.
When build for production, separate to two js files (or more), usually we will have one file called vendor.js for all libraries and app.js for just the code we made. This will leverage cache on browser as vendor.js won't change much each built.
I saw that there is a little information about using React for different pages, so I learned a lot from Gulp documentation and I found a lot of very small JS libraries MicroJS which could replace Dispatcher and EventEmmiter with just a 4-6 KB in size of course you need to do some manual work with them, but they saving about 20 times in JS file size.
Here is my Gulp file for generating minified react bundle for each page. I'm using Django for backend.
var gulp = require('gulp');
var uglify = require('gulp-uglify');
var htmlreplace = require('gulp-html-replace');
var source = require('vinyl-source-stream');
var browserify = require('gulp-browserify');
var reactify = require('reactify');
var babelify = require('babelify');
var streamify = require('gulp-streamify');
var fs = require("fs");
var rename = require("gulp-rename");
var path = {
APP_DIR: './apps/*.jsx',
OUT_DIR: './../static/apps',
};
process.env.NODE_ENV = 'development';
gulp.task('jsx', function() {
return gulp.src([path.APP_DIR])
.pipe(browserify({
insertGlobals : true,
debug : true,
transform: [reactify, babelify.configure({
presets: ["es2015", "react"]
})],
}))
.pipe(rename({
extname: ".min.js"
}))
.pipe(gulp.dest(path.OUT_DIR));
});
gulp.task('jsx_min', function() {
return gulp.src([path.APP_DIR])
.on('error', function (error) {
console.log(error);
})
.pipe(browserify({
insertGlobals : true,
debug : false,
transform: [reactify, babelify.configure({
presets: ["es2015", "react"]
})],
}))
.pipe(streamify(uglify().on('error', function (e) {
console.log(e);
})))
.pipe(rename({
extname: ".min.js"
}))
.pipe(gulp.dest(path.OUT_DIR));
});
gulp.task('build', ['jsx_min']);
gulp.task('default', ['jsx'], function () {
return gulp.watch([path.APP_DIR], ['jsx', function () {
var current_date = new Date();
var time = current_date.getHours() + ":" + current_date.getMinutes() + ":" + current_date.getSeconds();
console.log(time, " -> Rebuilding");
}]);
});
Now for each logical page I got about 40KB minified JS file for handling all JavaScript including AJAX functionality.
So I'm marking my question answered :)
Thanks for help.
In make it's possible to define custom targets that have no relevance to the actual code that they act upon, in the sense that they are language agnostic.
release_sortof:
#echo packaging release...
tar czf release.tar.gz file1 file2 file3
ls /dev/null
ls /dev/stderr
ls /dev/stdout
I know the example above is horrible, but the point I'm trying to illustrate is that the code in the release_sortof target doesn't depend on the fact that my project uses code written in C, for example; nor does it depend on me using Make built-ins such as foreach.
Is there a way to work with javascript/<INSERT-NAME>script files without using the ever insufficient plugins available for gulp? As in, could I lint my coffeescript with coffeelint by directly calling the coffeelint module:
var gulp = require('gulp')
, coffeelint = require('coffeelint')
;
gulp.task('lint', function() {
/* run coffeelint on source files */
});
Or can this only be done using plugins?
Another example would be to run arbitrary code like so:
var spawn = require('child_process').spawn;
gulp.task('blue', function() {
var child = spawn('ls');
/* do stuff with spawned child process */
});
I do this kind of thing for browserify using vinyl-source-stream - basically allowing you to use the library as it is, and not using gulp-* plugins.
var browserify = require('browserify'),
gulp = require('gulp'),
source = require('vinyl-source-stream'),
stringify = require('stringify'),
plumber = require('gulp-plumber'),
config = require('../config').scripts;
gulp.task('browserify', function () {
return browserify(config.app)
.transform(stringify(['.html']))
.bundle()
.pipe(plumber())
.pipe(source('bundle.js'))
.pipe(gulp.dest(config.dest));
});
Heres the npm - https://www.npmjs.com/package/vinyl-source-stream
Use conventional text streams at the start of your gulp or vinyl
pipelines, making for nicer interoperability with the existing npm
stream ecosystem.
Maybe that will help you?
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'])
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.