I am working on a simple gulpfile and noticed an issue with gulp.watch method. If I add a new file to an empty directory gulp.watch will not fire. However if there is at least one file in the directory all change events are detected. I could obviously restart my "watch" task every time there is an empty directory added with a new file or I add a file to an existing empty directory but that seems counter intuitive to the purpose of gulp.watch method.
To be clear watch does detect files that are added and deleted only after at least one file exists in that directory.
My question is wether or not this is a bug exclusive to me or if more people have experienced this. Also does anyone know of a current work around?
Here is my gulp task:
gulp.task('watch', () => {
var watcher = gulp.watch('src/styles/scss/*.scss', {cwd: './'}, ['styles']);
watcher.on('change', (event) => {
console.log(`File ${event.path} was ${event.type}, running tasks...`);
});
Current gulp version: 3.9.1
P.S. I also know this may be a limitation of the technology I just don't what to report a bug to the gulp team that isn't a bug.
Thanks!
Awesome! Thank you, Mark for getting me in the right direction. It is not a bug there is just a specific way you have to do it.
gulp.task('watch', () => {
var watcher = gulp.watch(['src/styles/scss/*.scss', 'src/styles/*],{cwd: './'}, ['styles']);
watcher.on('change', (event) => {
console.log(`File ${event.path} was ${event.type}, running tasks...`);
});
The trick is watching your parent directory for any changes. This will now detect file changes as well as added and deleted files in empty subdirectories.
Related
I have the following setup:
// watch for changes
gulp.task('watch', function () {
gulp.watch('./assets/**/*.less', ['compile-less']);
});
gulp.task("compile-less", () => {
return gulp.src('./assets/build-packages/*.less')
.pipe($.less({
paths: [ $.path.join(__dirname, 'less', 'includes') ]
}))
.pipe(gulp.dest(OutputPath)); // ./dist/styles/
});
So basically every time a developer changes something in a less file it runs the task 'compile-less'. The task 'compile-less' builds our package less files (including all the #imports). The first change in a random less file works, all the less files are being build. The second time it runs the task but my generated dist folder isn't updated when I change something to a less file that is imported. I'm wondering if the combination of the watch task and the compiling task somehow caches files. Because if I run the compile-less task manually it works everytime.
Does anyone had the same experience?
gulp-less version 4.0.0 has a strange caching issue.
Install gulp-less#3.5.0 and will solve the issue.
This will be fixed. Check out https://github.com/stevelacy/gulp-less/issues/283#ref-issue-306992692
So I have a task like so:
gulp.task('scripts', function() {
return gulp.src(['app/scripts/app.js', 'app/scripts/controllers/**/*.js', 'app/scripts/services/**/*.js', 'app/scripts/directives/**/*.js', 'app/scripts/libs/**/*.js' ])
.pipe(concat('external.min.js'))
.pipe(ngAnnotate())
.pipe(uglify())
.pipe(gulp.dest('app/scripts'))
.pipe(gulp.dest('dist/scripts'))
});
and I have a watch task:
gulp.task('watch', ['sass-dev', 'scripts'], function() {
gulp.watch('app/style/sass/**/*.scss', ['sass-dev']);
gulp.watch('app/scripts/**/*.js', ['scripts']);
});
All works well, except whenever I save a JS file, "scripts" runs multiple times. I'm assuming the problem lies with the gulp.src and it looking at multiple files, but I'm not sure.
This is no big deal (to me), but sometimes I'll swap over to the command line and the task is running infinitely. It just keeps getting called over and over again.
If you haven't already guessed, I'm running Angular, which is why app.js is first and I have ngAnnotate.
Can someone shed some light on why the script runs continuously sometimes?
I guess the problem is .pipe(gulp.dest('app/scripts')). You're doing some stuff (uglify and angular stuff) with your scripts and then you place them in the same folder you're watching. So the scripts task will launch again and again and again.
You should remove this line and only place your distribution scripts in your distribution folder and leave your app files untouched.
Im trying to use gulp and jscs to prevent code smell. I also want to use watch so that this happens when ever a change is made. The problem I'm running into is jscs is modify the source file that is being watched. This causes gulp to go into an infinite loop of jscs modifying the file and then watch seeing the change and firing off jscs again and again and again ...
const gulp = require('gulp');
gulp.task('lint', function() {
return gulp.src('/src/**/*.js')
.pipe(jscs({
fix: true
}))
.pipe(jscs.reporter())
.pipe(gulp.dest('/src'));
});
gulp.task('watch', function() {
gulp.watch('/src/**/*.js', ['lint']);
});
It's generally a bad idea to override source files from a gulp task. Any Editors/IDEs where those files are open might or might not handle that gracefully. It's generally better to write the files into a separate dist folder.
That being said here's two possible solutions:
Solution 1
You need to stop the gulp-jscs plugin from running a second time and writing the files again, thus preventing the infinite loop you're running into. To achieve this all you have to do is add gulp-cached to your lint task:
var cache = require('gulp-cached');
gulp.task('lint', function() {
return gulp.src('/src/**/*.js')
.pipe(cache('lint'))
.pipe(jscs({
fix: true
}))
.pipe(cache('lint'))
.pipe(jscs.reporter())
.pipe(gulp.dest('/src'));
});
The first cache() makes sure that only files on disk that have changed since the last invocation of lint are passed through. The second cache() makes sure that only files that have actually been fixed by jscs() are written to disk in the first place.
The downside of this solution is that the lint task is still being executed twice. This isn't a big deal since during the second run the files aren't actually being linted. gulp-cache prevents that from happening. But if you absolutely want to make sure that lint is run only once there's another way.
Solution 2
First you should use the gulp-watch plugin instead of the built-in gulp.watch() (that's because it uses the superior chokidar library instead of gaze).
Then you can write yourself a simple pausableWatch() function and use that in your watch task:
var watch = require('gulp-watch');
function pausableWatch(watchedFiles, tasks) {
var watcher = watch(watchedFiles, function() {
watcher.close();
gulp.start(tasks, function() {
pausableWatch(watchedFiles, tasks);
});
});
}
gulp.task('watch', function() {
pausableWatch('/src/**/*.js', ['lint']);
});
In the above the watcher is stopped before the lint task starts. Any .js files written during the lint task will therefore not trigger the watcher. After the lint task has finished, the watcher is started up again.
The downside of this solution is that if you save a .js file while the lint task is being executed that change will not be picked up by the watcher (since it has been stopped). You have to save the .js file after the lint task has finished (when the watcher has been started again).
I am a new gulp user. I am in a situation here. I want to run a process if any change happens in any of the ".js" files changes.
what i did was
gulp.task('watch', function() {
watch('./src/**/*.js', batch(function() {
gulp.start('js-process');
}));
});
js-process is the task which i have to run.
To start the process.I did ,
gulp.task('default', ['js-process','css-process', 'copy-otherfiles','watch']);
My problem is , its working only once, At first time if any change occurs it works fine, But for further any change in any .js files, the watch doesn't work
Any Ideas? Thanks
I'm using gulp-watch plugin and would like to copy newly added files in the source to the target destination.
watch({glob:SOURCE + '/**/*.js'})
.pipe(plumber())
.pipe(gulp.dest(DESTINATION));
Every time a new file is added into the SOURCE directory I get "Bus error: 10" and the watch breaks without copying the newly added file.
Please you this syntax for adding new files in gulp
gulp.task('task_name', function() {
return watch({
glob: SOURCE
}, function(files) {
return files.pipe(plumber()).pipe(jade()).pipe(gulp.dest(DESTINATION));
});
});
gulp.watch doesn't create a source stream, it triggers on file changes and calls tasks.
Try creating a simple move task alongside a watch task, then trigger the move from the watch. Something like this:
gulp.task('move-js', function() {
gulp.src('./js-src/**/*.js')
.pipe(gulp.dest('./js-dest'));
});
gulp.task('watch-js', ['move-js'], function() {
gulp.watch('./js-src/**/*.js', ['move-js']);
});
Note that the watch-js task has move-js as a dependency, this will call the move task whenever the watch is invoked, rather than waiting for something in the watched directory to change.
I repeated the glob for clarity, but that should probably be stored in a variable.