Chain Gulp glob to browserify transform - javascript

I have a project with a few relatively disjoint pages, each including their own entry point script. These scripts require a number of others using commonjs syntax, and need to be transformed by 6to5 and bundled by browserify.
I would like to set up a gulp task that captures all the files matching a pattern and passes them on to the bundler, but I'm not sure how to pass files from gulp.src to browserify(filename).
My gulpfile looks like:
var gulp = require("gulp");
var browserify = require("browserify");
var to5browserify = require("6to5-browserify");
var source = require("vinyl-source-stream");
var BUNDLES = [
"build.js",
"export.js",
"main.js"
];
gulp.task("bundle", function () {
/* Old version, using glob:
return gulp.src("src/** /*.js")
.pipe(sixto5())
.pipe(gulp.dest("dist"));
*/
// New version, using array:
return BUNDLES.map(function (bundle) {
return browserify("./src/" + bundle, {debug: true})
.transform(to5browserify)
.bundle()
.pipe(source(bundle))
.pipe(gulp.dest("./dist"));
});
});
gulp.task("scripts", ["bundle"]);
gulp.task("html", function () {
return gulp.src("src/**/*.html")
.pipe(gulp.dest("dist"));
});
gulp.task("styles", function () {
return gulp.src("src/**/*.css")
.pipe(gulp.dest("dist"));
});
gulp.task("default", ["scripts", "html", "styles"]);
This seems to work, but isn't maintainable: I'll be adding more scripts relatively soon, and don't want to add them to the array every time.
I've tried using gulp.src(glob).pipe within the browserify call and piping after calling (shown here), and gulp.src(glob).map (method doesn't exist).
How can you chain gulp.src with a name-based transformer like browserify?

Use through2 to make a one-off custom plugin stream that does all of the dirty work.
Unfortanately vinyl-transform and vinyl-source-stream and the solutions that go along with those have flaws so we have to go for something custom.
var gulp = require('gulp');
var through = require('through2');
var browserify = require('browserify');
gulp.task('bundle', function() {
var browserified = function() {
return through.obj(function(chunk, enc, callback) {
if(chunk.isBuffer()) {
var b = browserify(chunk.path);
// Any custom browserify stuff should go here
//.transform(to5browserify);
chunk.contents = b.bundle();
this.push(chunk);
}
callback();
});
};
return gulp.src(['./src/**/*.js'])
.pipe(browserified())
.pipe(gulp.dest('dest'));
});

You can specify globs in your BUNDLES array as well as exclude any files:
var BUNDLES = [
"app/**/*.js",
"export.js",
"app/modules/**/*.js",
"!app/modules/excluded/*.js"
];

Related

How to make Gulp watch monitoring files?

I have a simple gulpfile.js, that defines only two tasks, buildLess and watchFiles:
var gulp = require('gulp');
var less = require('gulp-less');
var watch = require('gulp-watch');
var plumber = require('gulp-plumber');
var filter = require('gulp-filter');
function buildLess(done) {
const fileFilter = filter(['**/*', '!**/mixins.less', '!**/variables.less']);
gulp.src('./public/less/*.less')
.pipe(fileFilter)
.pipe(plumber())
.pipe(less())
.pipe(gulp.dest('./public/css/'))
;
done();
};
function watchFiles() {
gulp.watch(['public/less/*.less'], gulp.series('build-less'));
// gulp.watch(['./public/less/*.less'], gulp.series(buildLess));
};
gulp.task('build-less', buildLess);
gulp.task('watch-files', watchFiles);
The first one ($ gulp build-less) is working fine. The watchFiles ($ gulp watch-files) can be started and doesn't cause any errors, but changes on the public/less/style.less are ignored.
What is wrong at this gulpfile.js and how to get the watch-files task working?
The gulp.series API allows you to pass a string of a previously registered task. In your code, you haven't registered build-less yet.
var gulp = require('gulp');
var less = require('gulp-less');
var watch = require('gulp-watch');
var plumber = require('gulp-plumber');
var filter = require('gulp-filter');
function buildLess(done) {
const fileFilter = filter(['**/*', '!**/mixins.less', '!**/variables.less']);
gulp.src('./public/less/*.less')
.pipe(fileFilter)
.pipe(plumber())
.pipe(less())
.pipe(gulp.dest('./public/css/'))
;
done();
};
gulp.task('build-less', buildLess);
function watchFiles() {
gulp.watch(['public/less/*.less'], gulp.series('build-less'));
// gulp.watch(['./public/less/*.less'], gulp.series(buildLess));
};
gulp.task('watch-files', watchFiles);
I would note that Gulp does not recommend using the gulp.task API anymore to register tasks, but instead to use exports.
Secondly, you don't need gulp-watch, as gulp now comes with its own gulp.watch method (which you are already using).
Lastly, you should make sure to your correctly signaling async completion in your buildLess function. Below, I've changed that function to return a Stream, rather than calling a done() callback since as you have it written, you have a race condition where done() may be called before the Less compilation has finished.
var gulp = require('gulp');
var less = require('gulp-less');
var plumber = require('gulp-plumber');
var filter = require('gulp-filter');
function buildLess() {
const fileFilter = filter(['**/*', '!**/mixins.less', '!**/variables.less']);
return gulp
.src('./public/less/*.less')
.pipe(fileFilter)
.pipe(plumber())
.pipe(less())
.pipe(gulp.dest('./public/css/'));
}
exports['build-less'] = buildLess;
function watchFiles() {
gulp.watch(['public/less/*.less'], buildLess);
}
exports['watch-files'] = watchFiles;
Overall, I'd go through Gulp's documentation. They recently updated their website, and updated their documentation along with it. Going through that might clear up some other questions you may be having.

gulp-load-plugins.sourcemaps.init() TypeError: Cannot read property 'init' of undefined

I'm trying to adapt a gulp file to my purposes and I'm running into issues. I only care about one task:
gulp.task('js:browser', function () {
return mergeStream.apply(null,
Object.keys(jsBundles).map(function(key) {
return bundle(jsBundles[key], key);
})
);
});
It is using browserify to condense my bundle into a usable single file. It uses these two methods and this object:
function createBundle(src) {
//if the source is not an array, make it one
if (!src.push) {
src = [src];
}
var customOpts = {
entries: src,
debug: true
};
var opts = assign({}, watchify.args, customOpts);
var b = watchify(browserify(opts));
b.transform(babelify.configure({
stage: 1
}));
b.transform(hbsfy);
b.on('log', plugins.util.log);
return b;
}
function bundle(b, outputPath) {
var splitPath = outputPath.split('/');
var outputFile = splitPath[splitPath.length - 1];
var outputDir = splitPath.slice(0, -1).join('/');
console.log(outputFile);
console.log(plugins);
return b.bundle()
// log errors if they happen
.on('error', plugins.util.log.bind(plugins.util, 'Browserify Error'))
.pipe(source(outputFile))
// optional, remove if you don't need to buffer file contents
.pipe(buffer())
// optional, remove if you dont want sourcemaps
.pipe(plugins.sourcemaps.init({loadMaps: true})) // loads map from browserify file
// Add transformation tasks to the pipeline here.
.pipe(plugins.sourcemaps.write('./')) // writes .map file
.pipe(gulp.dest('build/public/' + outputDir));
}
var jsBundles = {
'js/polyfills/promise.js': createBundle('./public/js/polyfills/promise.js'),
'js/polyfills/url.js': createBundle('./public/js/polyfills/url.js'),
'js/settings.js': createBundle('./public/js/settings/index.js'),
'js/main.js': createBundle('./public/js/main/index.js'),
'js/remote-executor.js': createBundle('./public/js/remote-executor/index.js'),
'js/idb-test.js': createBundle('./public/js/idb-test/index.js'),
'sw.js': createBundle(['./public/js/sw/index.js', './public/js/sw/preroll/index.js'])
};
When I run the gulp task js:bower I get the following error coming from the the .pipe(plugins.sourcemaps.init({loadMaps: true})) expression:
TypeError: Cannot read property 'init' of undefined
I know that the lines are optional and I can just comment them out, but I do want them. When I run the code in the example file it works properly, when I run it in my gulp file it gives me the error. Any suggestions on what I might be missing? Thanks!
gulp-load-plugins analyzes the contents of your package.json file to find out which Gulp plugins you have installed. Make sure that gulp-sourcemaps is among the "devDependencies" defined there. If not run
npm install --save-dev gulp-sourcemaps
There's a small chance that your problem is related to lazy loading the sourcemaps plugin. If the above doesn't help try requiring gulp-load-plugins like this:
var plugins = require('gulp-load-plugins')({lazy:false});

How do you use browserify in a Gulp task?

I'm pretty new to Gulp, but by following this tutorial I set up a Gulp task that is meant to browserify javascript files in a particular directory and pipe them to a different directory - pretty simple. I've looked a few other tutorials, but this method seemed to be the most concise. Here is my code:
var gulp = require('gulp');
var browserify = require('browserify');
var transform = require('vinyl-transform');
gulp.task('js', function() {
var browserified = transform(function(filename) {
return browserify(filename).bundle();
});
return gulp.src(['./public/js/src/**/*.js'])
.pipe(browserified)
.pipe(gulp.dest('./public/js/dist'));
});
The above code is very similar to many other implementations of this sort I've seen, but when I try running it with gulp js, it produces the following error:
[15:47:13] Using gulp file
~/development/launchpad/workshop/gulpfile.js
[15:47:13] Starting 'js'...
_stream_readable.js:540
var ret = dest.write(chunk);
^
TypeError: undefined is not a function
at Producer.ondata (_stream_readable.js:540:20)
at Producer.emit (events.js:107:17)
at Producer.Readable.read (_stream_readable.js:373:10)
at flow (_stream_readable.js:750:26)
at resume_ (_stream_readable.js:730:3)
at _stream_readable.js:717:7
at process._tickCallback (node.js:355:11)
Does anyone know what might cause this error?
(As a side note, I'd like to look at the files from the stack trace to try to figure out what is going on here, but searching for _stream_readable.js in Spotlight yields about 20 files of that name, all seemingly Node modules. Is there a way to determine the full path of a file in a stack trace?)
var browserify = require('browserify');
var gulp = require('gulp');
var source = require('vinyl-source-stream');
gulp.task('browserify', function() {
return browserify('lib/front/app.js')
.bundle()
//Pass desired output filename to vinyl-source-stream
.pipe(source('bundle.js'))
// Start piping stream to tasks!
.pipe(gulp.dest('public/build/'));
});
If you want browserify to work with gulp. dest and create a file where we specify it via .pipe (gulp.dest ('src/js')),
then you need to download vinyl-source-stream and throw it in .pipe(source('bundle.js')),
but actually in browserify, namely the bundle method accepts callback and neither dest nor source is needed
browserify({
entries: jsFile,
basedir: "src/js/dev",
debug: true,
})
.transform(babelify, {
presets: ['#babel/preset-env'],
})
.bundle((err, buffer) => {
let event = new EventEmitter();
if (err) {
event.emit('error',err)
}
else {
let data = minify(buffer.toString(), {}).code;
fs.createWriteStream('./src/js/bundle.js').write(data)
console.dir(222);
bs.reload()
}
})
Unfortunately, this is an issue with browserify/gulp, and there's nothing that vinyl-transform can do. The solution is to use vinyl-source-stream and vinyl-buffer:
var gulp = require('gulp');
var browserify = require('browserify');
var source = require('vinyl-source-stream');
var glob = require('node-glob');
gulp.task('browserify', function (cb) {
glob('./src/**/*.js', {}, function (err, files) {
var b = browserify();
files.forEach(function (file) {
b.add(file);
});
b.bundle().
.pipe(source('output.js'))
.pipe(gulp.dest('./dist'));
cb();
})
});
More information here.

es.merge doesn't merge my streams

I have the following gulp task:
var es = require('event-stream'),
concat = require('gulp-concat'),
templateCache = require('gulp-angular-templatecache');
var scripts = gulp.src(paths.js + '/**/*.js'),
templates = gulp.src(paths.templates + '/**/*.html')
.pipe(templateCache('templates.js', {
module: 'spot.im.core',
standalone: false
}));
es.merge(templates, scripts)
.pipe(concat('all.js'))
.pipe(gulp.dest(paths.dist))
When running this, I don't get the templates inside all.js. But, when I am doing:
es.merge(templates)
.pipe(concat('all.js'))
.pipe(gulp.dest(paths.dist))
all.js contains the templates.
What am I doing wrong?
I think you forgot to define a task here; and, you can replace event-stream with merge-stream instead. I've had no trouble running merge-stream in many gulp tasks.
var merge = require('merge-stream'),
concat = require('gulp-concat'),
templateCache = require('gulp-angular-templatecache');
gulp.task('default', function () {
var scripts = gulp.src(paths.js + '/**/*.js');
var templates = gulp.src(paths.templates + '/**/*.html')
.pipe(templateCache('templates.js', {
module: 'spot.im.core',
standalone: false
}));
return merge(templates, scripts)
.pipe(concat('all.js'))
.pipe(gulp.dest(paths.dist));
});
Run on the command line with gulp.

Iterating over directories with Gulp?

I'm new to gulp, but I'm wondering if its possible to iterate through directories in a gulp task.
Here's what I mean, I know a lot of the tutorials / demos show processing a bunch of JavaScript files using something like "**/*.js" and then they compile it into a single JavaScript file. But I want to iterate over a set of directories, and compile each directory into it's own JS file.
For instance, I have a file structure like:
/js/feature1/something.js
/js/feature1/else.js
/js/feature1/foo/bar.js
/js/feature1/foo/bar2.js
/js/feature2/another-thing.js
/js/feature2/yet-again.js
...And I want two files: /js/feature1/feature1.min.js and /js/feature2/feature2.min.js where the first contains the first 4 files and the second contains the last 2 files.
Is this possible, or am I going to have to manually add those directories to a manifest? It would be really nice to pragmatically iterate over all the directories within /js/.
Thanks for any help you can give me.
-Nate
Edit: It should be noted that I don't only have 2 directories, but I have many (maybe 10-20) so I don't really want to write a task for each directory. I want to handle each directory the same way: get all of the JS inside of it (and any sub-directories) and compile it down to a feature-based minified JS file.
There's an official recipe for this: Generating a file per folder
var fs = require('fs');
var path = require('path');
var merge = require('merge-stream');
var gulp = require('gulp');
var concat = require('gulp-concat');
var rename = require('gulp-rename');
var uglify = require('gulp-uglify');
var scriptsPath = 'src/scripts';
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
gulp.task('scripts', function() {
var folders = getFolders(scriptsPath);
var tasks = folders.map(function(folder) {
return gulp.src(path.join(scriptsPath, folder, '/**/*.js'))
// concat into foldername.js
.pipe(concat(folder + '.js'))
// write to output
.pipe(gulp.dest(scriptsPath))
// minify
.pipe(uglify())
// rename to folder.min.js
.pipe(rename(folder + '.min.js'))
// write to output again
.pipe(gulp.dest(scriptsPath));
});
// process all remaining files in scriptsPath root into main.js and main.min.js files
var root = gulp.src(path.join(scriptsPath, '/*.js'))
.pipe(concat('main.js'))
.pipe(gulp.dest(scriptsPath))
.pipe(uglify())
.pipe(rename('main.min.js'))
.pipe(gulp.dest(scriptsPath));
return merge(tasks, root);
});
You could use glob to get a list of directories and iterate over them, using gulp.src to create a separate pipeline for each feature. You can then return a promise which is resolved when all of your streams have ended.
var fs = require('fs');
var Q = require('q');
var gulp = require('gulp');
var glob = require('glob');
gulp.task('minify-features', function() {
var promises = [];
glob.sync('/js/features/*').forEach(function(filePath) {
if (fs.statSync(filePath).isDirectory()) {
var defer = Q.defer();
var pipeline = gulp.src(filePath + '/**/*.js')
.pipe(uglify())
.pipe(concat(path.basename(filePath) + '.min.js'))
.pipe(gulp.dest(filePath));
pipeline.on('end', function() {
defer.resolve();
});
promises.push(defer.promise);
}
});
return Q.all(promises);
});
I am trying myself to get how streams work in node.
I made a simple example for you, on how to make a stream to filter folders and start a new given stream for them.
'use strict';
var gulp = require('gulp'),
es = require('event-stream'),
log = require('consologger');
// make a simple 'stream' that prints the path of whatever file it gets into
var printFileNames = function(){
return es.map(function(data, cb){
log.data(data.path);
cb(null, data);
});
};
// make a stream that identifies if the given 'file' is a directory, and if so
// it pipelines it with the stream given
var forEachFolder = function(stream){
return es.map(function(data, cb){
if(data.isDirectory()){
var pathToPass = data.path+'/*.*'; // change it to *.js if you want only js files for example
log.info('Piping files found in '+pathToPass);
if(stream !== undefined){
gulp.src([pathToPass])
.pipe(stream());
}
}
cb(null, data);
});
};
// let's make a dummy task to test our streams
gulp.task('dummy', function(){
// load some folder with some subfolders inside
gulp.src('js/*')
.pipe(forEachFolder(printFileNames));
// we should see all the file paths printed in the terminal
});
So in your case, you can make a stream with whatever you want to make with the files in a folder ( like minify them and concatenate them ) and then pass an instance of this stream to the forEachFolder stream I made. Like I do with the printFileNames custom stream.
Give it a try and let me know if it works for you.
First, install gulp-concat & gulp-uglify.
$ npm install gulp-concat
$ npm install gulp-uglify
Next, do something like:
//task for building feature1
gulp.task('minify-feature1', function() {
return gulp.src('/js/feature1/*')
.pipe(uglify()) //minify feature1 stuff
.pipe(concat('feature1.min.js')) //concat into single file
.pipe(gulp.dest('/js/feature1')); //output to dir
});
//task for building feature2
gulp.task('minify-feature2', function() { //do the same for feature2
return gulp.src('/js/feature2/*')
.pipe(uglify())
.pipe(concat('feature2.min.js'))
.pipe(gulp.dest('/js/feature2'));
});
//generic task for minifying features
gulp.task('minify-features', ['minify-feature1', 'minify-feature2']);
Now, all you have to do to minify everything from the CLI is:
$ gulp minify-features
I had trouble with the gulp recipe, perhaps because I'm using gulp 4 and/or because I did not want to merge all my folders' output anyway.
I adapted the recipe to generate (but not run) an anonymous function per folder and return the array of functions to enable them to be processed by gulp.parallel - in a way where the number of functions I would generate would be variable. The keys to this approach are:
Each generated function needs to be a function or composition (not a stream). In my case, each generated function was a series composition because I do lots of things when building each module folder.
The array of functions needs to passed into my build task using javascript apply() since every member of the array needs to be turned into an argument to gulp.parallel in my case.
Excerpts from my function that generates the array of functions:
function getModuleFunctions() {
//Get list of folders as per recipe above - in my case an array named modules
//For each module return a function or composition (gulp.series in this case).
return modules.map(function (m) {
var moduleDest = env.folder + 'modules/' + m;
return gulp.series(
//Illustrative functions... all must return a stream or call callback but you can have as many functions or compositions (gulp.series or gulp.parallel) as desired
function () {
return gulp.src('modules/' + m + '/img/*', { buffer: false })
.pipe(gulp.dest(moduleDest + '/img'));
},
function (done) {
console.log('In my function');
done();
}
);
});
}
//Illustrative build task, running two named tasks then processing all modules generated above in parallel as dynamic arguments to gulp.parallel, the gulp 4 way
gulp.task('build', gulp.series('clean', 'test', gulp.parallel.apply(gulp.parallel, getModuleFunctions())));
`

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