Develop Reference

Using Gulp to Minify and Auto Update a CSS File

I have a Gulp task that minifies my CSS in one folder and then pipes it to another folder.
const gulp = require("gulp");
const cleanCSS = require("gulp-clean-css");
gulp.task("minify-css", () => {
return (
gulp
.src("./css/**/*.css")
.pipe(cleanCSS())
.pipe(gulp.dest("minified"))
);
});
The command gulp minify-css works perfectly. I don't want to have to continually type that command in the terminal tho.
I want the code below to watch my CSS file and when it changes I want the minify-css task to run and update my minified file but it doesn't work:
gulp.task("default", function(evt) {
gulp.watch("./css/**/*.css", function(evt) {
gulp.task("minify-css");
});
});
Any ideas on why this doesn't work? Thank you in advance!

The problem lies in the area where you are calling gulp.task('minify-css') inside the gulp.watch callback function. That code does not actually invoke the minify-css task, but rather spawns an anonymous task as pointed by you in your logs. Instead gulp.watch should invoke a function which internally performs the minify-css job.
The other issue is probably the syntax changes that happened in gulp-4. They are not many but can be confusing.
I've managed to fix the issue and it works. Here is the updated code for gulpfile.js below:
const gulp = require("gulp");
const cleanCSS = require("gulp-clean-css");
function minifyCSS() {
return (
gulp
.src("./css/*.css")
.pipe(cleanCSS())
.pipe(gulp.dest("minified"))
);
}
gulp.task("minify-css", minifyCSS);
gulp.task("watch", () => {
gulp.watch("./css/*.css", minifyCSS);
});
gulp.task('default', gulp.series('minify-css', 'watch'));
Hope this helps.

I'm using Gulp and failing to produce the final development script for production.

So I'm having a slight problem with producing production ready scripts for my project. I'm using gulp to concatenate and minify my css and js, and while the css is working fine the gulp js function isn't generating my final file. Please refer to my code below:
gulp.task('js', function() {
return gulp.src([source + 'js/app/**/*.js'])
.pipe(concat('development.js'))
.pipe(gulp.dest(source + 'js'))
.pipe(rename({
basename: 'production',
suffix: '-min',
}))
.pipe(uglify())
.pipe(gulp.dest(source + 'js/'))
.pipe(notify({ message: 'Scripts task complete', onLast: true }));
});
If anyone has encountered a similar problem or has any tips it would be much appreciated :)

There is nothing wrong with your gulpfile. I tested it and it works perfectly.
The only thing I can guess is that your source is not set correctly. Did you forget the trailing slash '/' ?
I would suggest 2 things to figure it out. Include node path library to check where source is actually pointing to like this:
var path = require('path');
// in gulp task ...
path.resolve(path.resolve(source + 'js/app'));
Make sure it points where you think it does.
Secondly, you could use gulp-debug to establish that any files are found:
npm install gulp-debug
Then
var debug = require('gulp-debug');
// in gulp task ...
return gulp.src([source + 'js/app/**/*.js'])
.pipe(concat('development.js'))
.pipe(debug())
.pipe(gulp.dest(source + 'js'))
.pipe(debug())
// etc.
Good luck!

Based on additional infomation in the comments I realise you are generating JS files in a separate process ...
gulp is asynchronous by default. What this boils down to is that all functions try to run at the same time - if you want a specific order it must be by design. This is great because it's very fast but can be a headache to work with.
Problem
Here is what's basically happening:
// SOME TASK THAT SHOULD BE RUN FIRST
gulp.task('copy-vendor-files-to-tempfolder', function (done) {
// copy files to vendor folder
done()
})
// SOME TASKS THAT DEPEND ON FIRST TASK
gulp.task('complile-styles', function () { /* independent task */ })
gulp.task('concat-vendor-files', function () { /* concat files in vendor folder. depends on vendor files existing */ })
// GENERAL TASK WHICH STARTS OTHERS
gulp.task('ready', ['copy-vendor-files-to-tempfolder', 'compile-styles', 'concat-vendor-files])
When you try to run:
$ gulp ready
GULP TASK WILL FAIL! Folder is being created at the same time!!
NOWHERE TO COPY FILES!
Solution
There are many solutions but the following module has come in handy for me again and again:
npm install run-sequence
Then in your gulpfile.js:
var runSequence = require('run-sequence')
gulp.task('ready', function (done) {
runSequence(
'create-folders', // do this first,
[
'copy-css-files',
'copy-html-files'
], // do these AFTER but in parallel
done // callback when ready
)
})
This will guarantee the folder exists when you try to run the other functions.
In your specific case, you should make sure the task that concatenates the JS files is run after the task that copies them out of vendor.
Note: I'm leaving other answer because it contains useful help for debugging similar issues.
HTH!

Using Jasmine Ajax

I am writing some tests with Jasmine. I am running those tests via Gulp. I want to use the Jasmine Ajax plugin. However, I cannot figure out how to include it into my tests. Right now, I have the following:
tests.js
describe('MyApp', function() {
beforeEach(function() {
jasmine.Ajax.install();
});
it('should run an ajax request', function() {
// test ajax
});
});
Once again, I'm running this via Gulp. So, in my gulpfile.js I have the following:
gulpfile.js
var gulp = require('gulp');
var jasmine = require('gulp-jasmine');
gulp.task('test', function() {
return gulp
.src('tests/*.js')
.pipe(jasmine());
});
When I execute this, I get the following from the command-line:
TypeError: Cannot read property 'install' of undefined.
Its like Jasmine Ajax isn't getting loaded. Yet, I'm not sure how to load it. Can someone please help me solve this issue?
Thank you.

I haven't tested this myself since I don't have enough info to recreate your setup, but you could try to put the mock-ajax.js file in the helpers directory, that is where the default configuration of jasmine-npm, which is used by gulp-jasmine, looks for them.

Running existing task with gulp-watch

I've got some tasks already defined in gulpfile.js and I want to use gulp-watch plugin (to run tasks on new files). My question is, because I couldn't find anything, can I run my existing tasks while running watch (from plugin) function?
var gulp = require('gulp'),
watch = require('gulp-watch'),
...;
gulp.task('lint', function () {
return gulp.src(path.scripts)
.pipe(jshint())
.pipe(jshint.reporter(stylish));
});
gulp.task('watch', function () {
watch({ glob: 'app/**/*.js' }); // Run 'lint' task for those files
});
Because I don't want to include watch() task in every task I have. I would like to have only 1 task - watch, which will combine all "watches".
----- EDIT ----
(as I probably didn't quite get my point):
I need to run task from inside of gulp('watch') task. for example:
like I did it with gulp.watch:
gulp.task('watch', function () {
gulp.watch('files', ['task1', 'task2']);
});
I need to do the same but with gulp-watch plugin, something like (I know it wouldn't work):
var watch = require('gulp-watch');
gulp.task('watch', function () {
watch({ glob: 'files' }, ['task1', 'task2']);
});

I have also run into the problem of wanting to use gulp-watch (not gulp.watch), needing to use the callback form, and having trouble finding a suitable way to run a task in the callback.
My use case was that I wanted to watch all stylus files, but only process the main stylus file that includes all the others. Newer versions of gulp-watch may address this but I'm having problems with 4.3.x so I'm stuck on 4.2.5.
gulp.run is deprecated so I don't want to use that.
gulp.start works well, but is also advised against by the gulp author, contra.
The run-sequence plugin works well and lets you define a run
order, but it is a self-proclaimed hack:
https://www.npmjs.com/package/run-sequence
Contra suggest writing plain old functions and calling
those. This is a new idea to me, but I think the example below captures the idea. https://github.com/gulpjs/gulp/issues/505
Take your pick.
var gulp = require('gulp'),
watch = require('gulp-watch'), // not gulp.watch
runSequence = require('run-sequence');
// plain old js function
var runStylus = function() {
return gulp.src('index.styl')
.pipe(...) // process single file
}
gulp.task('stylus', runStylus);
gulp.task('watch', function() {
// watch many files
watch('*.styl', function() {
runSequence('stylus');
OR
gulp.start('stylus');
OR
runStylus();
});
});
All of these are working for me without warnings, but I'm still unsure about getting the "done" callback from the 4.2.x version of gulp-watch.

You will most likely want to run specific tasks related to the files you are watching -
gulp.task('watch',['lint'], function () {
gulp.watch('app/**/*.js' , ['lint']);
});
You can also use the ['lint'] portion to run any required tasks when watch first gets called, or utilize the tasks to run async with
gulp.task('default', ['lint','watch'])

You can just call one task, that then includes both task
gulp.task('default', ['lint','watch'])
so here you would just call 'gulp'

gulp.task('watch', function() {
watch(files, function() {
gulp.run(['task1', 'task2']);
});
});
work fine, except a warning

gulp task get error: EISDIR, read?

gulp.task('dev', function () {
watch(['public/**', '!public/index.html'], function () {
del.sync(['web/**', '!web/.gitkeep', '!web/index.html', '!web/js/**']);
del.sync(['cordova/www/**', '!cordova/www/.gitkeep', '!cordova/www/index.html', '!cordova/www/js/**']);
gulp.src(['public/**', '!public/index.html', '!public/js/**'])
.pipe(gulp.dest('web/'))
.pipe(gulp.dest('cordova/www/'));
});
});
I'm using gulp setting up my development workflow. What I want to do is whenever any change happen in the public/, I want to wipe out everything in another two directory: web/ and cordova/www/, and write everything from public/ to web/ and cordova/www/.
I don't know what I'm doing wrong here. The dev gulp task job keeps throwing the error: EISDIR, read!