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())
Related
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 am using gulp for creating some css from less and have watch function. Everything working ok when there is no errors in less files, watch is calling less function and compile css. But when i have errors in less files, watch just breaks say where is error stop. When i fix error in less file, watch does not work anymore. I have to start it again, is it possible to see if there is error and just continue watching for compiling, here is my gulp.js
// Less to CSS task
var parentPath = './content/css/';
var sourceLess = parentPath;
var targetCss = parentPath;
gulp.task('less', function () {
return gulp.src([sourceLess + 'styles.less'])
.pipe(less({ compress: true }).on('error', gutil.log))
.pipe(autoprefixer('last 10 versions', 'ie 9'))
.pipe(minifyCSS({ keepBreaks: false }))
.pipe(gulp.dest(targetCss))
.pipe(notify('Less Compiled, Prefixed and Minified'));
});
gulp.task('watch', function () {
gulp.watch([sourceLess + 'styles.less'], ['less']); // Watch all the .less files, then run the less task
});
In gulp-less plugin documentation says that it doesn't have a built-in way to fail the task and keep the watcher active. But it says that you can't do it with stream-combiner2.
You can see the example here taked from the official gulp github repo.
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'])
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 using Gulp for watch and sass complier. When I start "watch" first time then "sass" complier runs and its create the css files as per given path. However when I change the .scss files then it doesn't call "sass" complier again. Following is is my these two tasks and variables.
gulp.task('sass', function () {
gulp.src(config.sassPath)
.pipe(sass())
.pipe(gulp.dest(config.cssPath))
.pipe(livereload());
});
gulp.task('watch', false, function () {
livereload.listen(8189);
gulp.src(config.watchPaths)
.pipe(watch(config.watchPaths, function (event) {
gulp.start( 'sass', 'js-hint', 'server','test');
livereload();
}))
.pipe(livereload());
});
Following command i use to run "watch" task
gulp watch
I do see "watch" is reloading when I am changing the .scss file. Following is log for this.
[19:49:30] public/sass/html-controls.scss was changed
[19:49:30] /Users/dkuma204/Desktop/Dilip/Projects/OPEN/SourceCode/AWF/OPENApp/application/public/sass/html-controls.scss reloaded.
Not sure what I am missing here. Please help.
Why it is so complicated? Try this:
gulp.task('watch', false, function () {
livereload.listen(8189);
gulp.watch(config.watchPaths,['sass', 'js-hint', 'server', 'test'])
});
And your every task which requires livereload should have .pipe(livereload()) at the end.
You shouldn't use gulp start. Here is one of comment from github discussion:
gulp.start is undocumented on purpose because it can lead to
complicated build files and we don't want people using it