Is there a way to run a Gulp task on Webstorm launch?
Tried looking through webstorm settings but don't see anything, also Gulp integration is brand new to Webstorm 9.
gulp.task('watch', function () {
gulp.watch(paths.watch.css, function(){ compileLESS(paths.src.css, paths.dest.css, {name: 'Style'}) });
gulp.watch(paths.watch.cssBootstrap, function(){ compileLESS(paths.src.cssBootstrap ,paths.dest.cssBootstrap, {suffix: '.min', name: 'Bootstrap'}) });
gulp.watch(paths.watch.scripts, ['scripts']);
gulp.watch(paths.watch.libraries, ['libraries']);
});
Just want a way to run Gulp 'watch' (if the project has gulp and a task called watch) when ever i start Webstorm.
As of WebStorm 11, this is possible.
Settings > Tools > Startup Tasks
Hit the green plus icon to Add New Configuration
Choose Gulp.js
Choose your tasks that you want to run from the Tasks: dropdown
Done!
No:( WebStorm (as well as other JetBrains IDEs) doesn't support 'on launch' actions - see https://youtrack.jetbrains.com/issue/WEB-11818#comment=27-787220
Best thing to do is set your gulp task (which sets up the watchers) as a 'before launch' task of your main configuration.
Works great for me. I need to start my server before I can start developing, and when I do, I know this will start my watchers and do any necessary processing for the first time.
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
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.
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.
I have a gulp task that performs some pretty common tasks - it runs jshint to validate my code, then concats and minimizes the files and outputs them into single .min.js files.
The task (appears) to execute flawlessly when I run it manually. But the second I try to use it in a $gulp.watch it no longer outputs my file (it still executes and executes jshint though).
The code in my task:
gulp.src(path.join(workingPath, folder, '/*.js'))
.pipe(jshint())
.pipe(jshint.reporter(stylish))
.pipe(jshint.reporter('fail')) //stop build if errors found
.on('error', function() {
console.log("Please review and correct jshint errors.");
this.end();
})
.pipe(order([ //order files before concat - ensure module definitions are first
"*.module.js",
"*.js"
]))
.pipe(concat(filename+'.js'))
.pipe(gulp.dest(destinationPath)) //full combined version
.pipe(uglify())
.pipe(rename(filename+'.min.js'))
.pipe(gulp.dest(destinationPath)) //minified combined version
.on('error',function() {
console.log("An error occurred during Gulp processing.");
this.end();
});
My gulp watch (the task is named 'components'):
gulp.watch(componentsBasePath+"/**/*.js",['components']);
One thing that I've noticed though is at the end of the manual run I see "Process finished with exit code..". And if I kill my gulp.watch it outputs "Process finished with exit code.." - then it DOES creates the output files!
My goal is to have my 'components' task create those output files every time it is triggered by the watch - not just when I kill the watch.
Thank you!
Cliff
Ok so my hacky way to fix the problem with jetbrains (im using phpstorm), you gotta understand 2 things.
gulp watchers act on file save.
jetbrains will not auto update the project files (as you have found out it uses a cache).
To get around this problem i created a macro called saveSync which does the following actions:
Save all
Synchronize
Synchronize
Synchronize
Why did i synchronize 3 times? Because gulp takes a few seconds to finish tasks (compiling, etc) and if you update before they finish obviously the project view doesn't get update properly. I haven't figured out a way to insert a time delay into the macro itself.
After i created the macro, i just rebound ctrl + s from save all to the macro, and it worked.
If there is a 'cleaner' way of doing this i have yet to discover it.
Ran this by someone else and he found the cause of the issue. Though - it's not Gulp related at all it turns out.
The IDE I was using updated the folder and file structure instantly when I manually ran my 'components' task, however it did not do the same when I ran the gulp.watch task. I am happy to report though that the files were being created successfully, they just never appeared in the IDE until I killed the task.
I have the following code fragment in my gulpfile.
gulp.task('static', function() {
return gulp.src(['./src/**', '!./src/js/**', '!./src/js/', '!./src/scss/', '!./src/scss/**'])
.pipe(gulp.dest(outputDir + '/'))
});
gulp.task('watch', function() {
gulp.watch(['./src/**', '!./src/js/**', '!./src/js/', '!./src/scss/', '!./src/scss/**'], ['static']);
});
gulp.task('dev', ['static']);
gulp.task('default', ['watch', 'dev']);
If I run gulp dev, gulp watch or gulp static, everything works fine. However, if I run just gulp (default), it does the static task 5 times. Can anyone help me out with why this is happening?
P.S. The paths passed to watch are such because if I don't disclude the directories as separate paths, it seems to be copying the empty directories js and scss for some reason.
Probably because you're not returning the tasks, and you need them to be asyc.
See this: Gulp.js task, return on src?
and the docs (also linked in SO post above) https://github.com/gulpjs/gulp/blob/master/docs/API.md#async-task-support
Also, the dev task looks redundant in its current form - you may as well use the task static directly, unless you plan to bundle in more tasks with dev