I have a gulp task to uglify my JS:
gulp.task('uglify', ['eslint'], () => {
return gulp.src(jsDest + '/*.js')
.pipe(rename({suffix: '.min'}))
.pipe(stripDebug())
.pipe(uglify())
.pipe(gulp.dest(jsDest));
});
It works fine, but as I am using a wildcard to identify JS files, I end up with new files named name.min.min.js rather than existing minified files being overwritten.
A work around I have come up with is to have an additional task which cleans out these files before I uglify again:
gulp.task('cleanUglified', () => {
return del.sync('dist/js/*.min.js');
});
gulp.task('uglify', ['cleanUglified', 'eslint'], () => {
...
});
While this works fine, I'm sure there must be a way to have my task ignore anything named *.min* and in fact overwrite any that already exist.
Solved - I can ignore the already minified files and overwrite them by using this pattern to select the files to uglify:
return gulp.src(jsDest + '/*[^.min].js')
EDIT
Following the below comments, I have updated to:
return gulp.src(['*.js', '!*.min.js'])
Related
do I understand this wrong or is something not right here?
I have this piece of code
gulp.task("minifyScripts", function () {
return gulp.src("assets/scripts/*.js")
.pipe(uglify())
.pipe(rev())
.pipe(gulp.dest('assets/scripts/min'))
.pipe(rev.manifest())
.pipe(revDel())
.pipe(gulp.dest('assets/scripts'))
.pipe(livereload())
.pipe(browserSync.stream());
});
My understanding was that this should remove old .js file when one with new hash is created, but its not...
Do you have any idea? Thanks so much!
You need to specify either dest in the options, or base in the manifest options—unless you're writing everything to the root directory.
Try:
pipe(revDel({dest: "assets/scripts/min"})
Seems that rev-del has a number of users, including myself, who are unable to get the plug-in to delete the old static files after gulp-rev hashes them.
Switching over to gulp-rev-delete-original simply worked OOB.
In the OP's use case, the updated solution would be:
const revDel = require("gulp-rev-delete-original")
gulp.task("minifyScripts", function () {
return gulp.src("assets/scripts/*.js")
.pipe(uglify())
.pipe(rev())
.pipe(revDel()) // call just after rev()
.pipe(gulp.dest('assets/scripts/min'))
.pipe(rev.manifest())
//.pipe(revDel()) ==> move to just after rev()
.pipe(gulp.dest('assets/scripts'))
.pipe(livereload())
.pipe(browserSync.stream());
});
i have problem, when i run gulp watch -> run task styles:build, and all of my less files was recompile. How i can compile only changed file?
gulp.task('styles:build', function () {
return gulp.src(pathes.src.styles)
.pipe(changed(pathes.build.styles), {extension: '.css'})
.pipe(print(function(filepath) {
return "➔ file was changed: " + filepath;
}))
.pipe(plumber())
.pipe(less({
plugins: [autoprefix, cleanCSSPlugin],
paths: ['./', 'web/styles']
}))
.pipe(gulp.dest(pathes.build.styles))
});
gulp.task('watch', function() {
gulp.watch(pathes.src.styles, ['styles:build'])
});
You need to modify the line below to add a closing parenthsis:
.pipe(changed(pathes.build.styles, {extension: '.css'}))
Also as I cautioned the first time the task is run it probably will pass through all files.
i think i found solution
just install lessChanged = require('gulp-less-changed')
and include him before less pipe
.pipe(lessChanged())
.pipe(less())
I don't know why the server still stops whenever there's an error in my js files even though I have jshint in my gulpfile. I installed jshint and included it in my project because it reports errors in js files, but it's still failing. How can I fix this?
gulp.task('scripts', () => {
return gulp.src('assets/js/src/*.js')
.pipe(jshint())
.pipe(jshint.reporter('jshint-stylish', {beep: true}))
.pipe(concat('main.js'))
.pipe(gulp.dest('assets/js/build/'))
.pipe(uglify())
.pipe(gulp.dest('assets/js/'))
.pipe(browserSync.stream({stream: true}));
});
gulp-jshint does what you says it does: it reports errors in JavaScript files. Nothing more, nothing less. It doesn't prevent defective JavaScript files from reaching later pipe stages like uglify() (which throws up and thus stops your server if there's any error in a JavaScript file).
If you want to prevent defective JavaScript files from wrecking your server, you need to put all the jshint stuff into it's own task and make sure that task fails when any JavaScript file has an error:
gulp.task('jshint', () => {
return gulp.src('assets/js/src/*.js')
.pipe(jshint())
.pipe(jshint.reporter('jshint-stylish', {beep: true}))
.pipe(jshint.reporter('fail'))
});
Then you need to make your scripts task depend on that jshint task:
gulp.task('scripts', ['jshint'], () => {
return gulp.src('assets/js/src/*.js')
.pipe(concat('main.js'))
.pipe(gulp.dest('assets/js/build/'))
.pipe(uglify())
.pipe(gulp.dest('assets/js/'))
.pipe(browserSync.stream({stream: true}));
});
Now your scripts task will only run when the jshint task was successful. If any JavaScript file was defective jshint will output the error to the console while your server continues to run using the last good version of your JavaScript.
The simplest fix would be to use gulp-plumber to handle the error a little more gracefully:
var plumber = require("gulp-plumber");
gulp.task('scripts', () => {
return gulp.src('assets/js/src/*.js')
.pipe(plumber())
.pipe(jshint())
.pipe(jshint.reporter('jshint-stylish', {beep: true}))
.pipe(concat('main.js'))
.pipe(gulp.dest('assets/js/build/'))
.pipe(uglify())
.pipe(gulp.dest('assets/js/'))
.pipe(browserSync.stream({stream: true}));
});
Personally, I don't like that solution because it will prevent your minified file from being updated. Here's what I would recommend:
var jshintSuccess = function (file) {
return file.jshint.success;
}
gulp.task('scripts', () => {
return gulp.src('assets/js/src/*.js')
.pipe(sourcemaps.init())
.pipe(jshint())
.pipe(jshint.reporter('jshint-stylish', {
beep: true
}))
.pipe(gulpif(jshintSuccess, uglify()))
.pipe(concat('main.js'))
.pipe(sourcemaps.write('maps'))
.pipe(gulp.dest('assets/js/'))
.pipe(browserSync.stream({
stream: true
}));
});
First, notice that I'm not writing to multiple destinations. Instead, I'm using sourcemaps so that you don't need unminified code. Second, I'm using gulp-if to conditionally pipe your code through uglify based on the results of jshint. Code with errors will bypass uglify so that it still makes it into to your destination file.
Now, you can inspect and debug it with the developer tools.
Note: I recommend this for local development only. I wouldn't connect this to a continuous integration pipeline because you'll only want good code to make it into production. Either set up a different task for that or add another gulp-if condition to prevent broken code from building based on environment variables.
I'm new to using gulp and I think I have it setup correctly, but it does not seem to be doing what it should be doing.
My gulpfile.js has
gulp.task('compass', function() {
return gulp.src('sites/default/themes/lsl_theme/sass/**/*.scss')
.pipe(compass({
config_file: 'sites/default/themes/lsl_theme/config.rb',
css: 'css',
sass: 'scss'
}))
.pipe(gulp.dest('./sites/default/themes/lsl_theme/css'))
.pipe(notify({
message: 'Compass task complete.'
}))
.pipe(livereload());
});
with
gulp.task('scripts', function() {
return gulp.src([
'sites/default/themes/lsl_theme/js/**/*.js'
])
.pipe(plumber())
.pipe(concat('lsl.js'))
.pipe(gulp.dest('sites/default/themes/lsl_theme/js'))
// .pipe(stripDebug())
.pipe(uglify('lsl.js'))
.pipe(rename('lsl.min.js'))
.pipe(gulp.dest('sites/default/themes/lsl_theme/js'))
.pipe(sourcemaps.write())
.pipe(notify({
message: 'Scripts task complete.'
}))
.pipe(filesize())
.pipe(livereload());
});
and the watch function
gulp.task('watch', function() {
livereload.listen();
gulp.watch('./sites/default/themes/lsl_theme/js/**/*.js', ['scripts']);
gulp.watch('./sites/default/themes/lsl_theme/sass/**/*.scss', ['compass']);
});
when I run gulp, the result is
[16:14:36] Starting 'compass'...
[16:14:36] Starting 'scripts'...
[16:14:36] Starting 'watch'...
[16:14:37] Finished 'watch' after 89 ms
and no changes are registered.
for file structure, my gulpfile.js is in the root directory and the sass, css, and js are all in root/sites/default/themes/lsl_theme with the sass folder containing the folder 'components' full of partials.
My assumption is that you are on windows? Correct me if I'm wrong.
There is this problem that gulp-notify tends to break the gulp.watch functions. Try commenting out
// .pipe(notify({
// message: 'Scripts task complete.'
// }))
and see if the problem still exists.
If that does fix the issue, a solution from this thread may be helpful.
You can use the gulp-if
plugin in combination with
the os node module
to determine if you are on Windows, then exclude gulp-notify, like
so:
var _if = require('gulp-if');
//...
// From https://stackoverflow.com/questions/8683895/variable-to-detect-operating-system-in-node-scripts
var isWindows = /^win/.test(require('os').platform());
//...
// use like so:
.pipe(_if(!isWindows, notify('Coffeescript compile successful')))
It turns out that a large part of my issue was just simply being a rookie with Gulp. When I removed 'scripts' from my gulp watch it started working.
I then made the connection that it was watching the same directory that it was placing the new concatenated and minified js files in so it was putting the new file, checking that file, and looping over and over causing memory issues as well as not allowing 'compass' to run.
After creating a 'dest' folder to hold the new js everything started working just peachy.
I am trying to do automation to concat and uglify js in gulp.
Here is my gulpfile.js:
gulp.task('compressjs', function() {
gulp.src(['public/app/**/*.js','!public/app/**/*.min.js'])
.pipe(sourcemaps.init())
.pipe(wrap('(function(){"use strict"; <%= contents %>\n})();'))
.pipe(uglify())
.pipe(concat('all.js'))
.pipe(rename({
extname: '.min.js'
}))
.pipe(sourcemaps.write('.'))
.pipe(gulp.dest('public/app'));
})
Do you think it is needed to wrap every file with (function(){"use strict"; <%= contents %>\n})(); to avoid conflict when every file is being join together? Do you think my gulp task is good, or it can even better for performing it's task?
Wrapping every file in a closure really isn't necessary for most code. There are some bad libs out there that leak vars, but I'd suggest you deal with them on a case by case basis and, if possible, issue Pull Requests to fix the problem or just stop using them. Usually, they can't be fixed as simply as wrapping them in a function.
The task you have above won't properly pass all files to the uglify task - you will need to concatenate first. You also don't need to rename as you can specify the full name in concatenate.
Below is a well tested Gulp setup for doing exactly what you've asked:
gulp.task('javascript:vendor', function(callback) {
return gulp.src([
'./node_modules/jquery/dist/jquery.js',
'./node_modules/underscore/underscore.js',
'./node_modules/backbone/backbone.js'
])
.pipe(sourcemaps.init())
// getBundleName creates a cache busting name
.pipe(concat(getBundleName('vendor')))
.pipe(uglify())
.pipe(sourcemaps.write('./'))
.pipe(gulp.dest('./public/app'))
.on('error', handleErrors);
});