I need some help.
I'm looking for a way to generate (or update if the file already exists) a .css file that is a conversion by an .scss file. All of this when compiling.
Explaining this in a better way :
I'm writing some code, everything is ok and I decide to save. Perfect. ctrl+s and the app run perfectly. Nice. Now I've added a style.scss file somewhere (it doesn't really matter the path). How do I "tell" to the compiler that everytime he compile he also has to 'take' this .scss file, convert it in a .css file, and put it in a specific path?
Well, I found a way to do what I needed to do.
I've created my gulpfile.js in this way :
var gulp = require('gulp');
var sass = require('gulp-sass');
var watch = require('gulp-watch');
gulp.task('styles', function () {
gulp.src('src/app/sass/*.scss')
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('./css/'));
});
gulp.task('watch', function () {
gulp.watch('./sass/**/*.scss', ['styles']);
});
And added this command to package.json :
"try": "gulp watch && ng s"
the problem is that if in the cli I run the command npm run try it will never start my application, because the watch is an endless stream.
How can I have the watch and the app running both at the same time?
*Edit
Found the solution using concurrently
Related
Following code, tried with both ./ as src and dest. Gave security administration full rights to folders, just in case if there was any issue.
So whenever I change styles.css from the styles folder, the code runs good on gulp watch and detects change too. It does run styles command on file change too. But then no folders are created in my dest folder.
var gulp = require('gulp');
gulp.task('styles' , function() {
return gulp.src('/app/assets/styles/styles.css')
.pipe(gulp.dest('/app/styles.css'));
});
gulp.task('watch',function(){
gulp.watch('./app/assets/styles/styles.css',
function() {
gulp.start('styles');
});
});
gulp.dest() takes a folder only, not a file. It probably is creating a folder, it is just called styles.css! Look under that for your file styles.css.
You probably want gulp.dest('./app'). the file name will be retained automatically.
I would also simplify to the below. gulp.start is not really documented.
gulp.task('watch',function(){
gulp.watch('./app/assets/styles/styles.css', ['styles']);
});
gulp.task('styles' , function() {
// added the . before /app here and in dest
return gulp.src('./app/assets/styles/styles.css')
.pipe(gulp.dest('./app'));
});
Similar to other questions, in this very watered-down snippet, running the default gulp task (via npm start which runs gulp); this snippet creates an infinite loop running the scripts task over and over. Here is the gulpfile.js (literally the whole thing at the moment):
'use strict';
const gulp = require('gulp');
// COPY SCRIPTS TO BUILD FOLDER
gulp.task('scripts', function() {
return gulp.src('./scripts/**/*.js')
.pipe(gulp.dest('build/scripts'))
});
// WATCH FILES
gulp.task('watch', function() {
gulp.watch('./scripts/**/*.js', ['scripts']);
});
// DEFAULT
gulp.task('default', ['watch']);
The extra odd thing is that whether the build folder is built anew or not, the scripts task will be executed immediately after calling npm start! And the loop begins.
In case you're curious, here is the pasted (and only) scripts object in my package.json:
"scripts": {
"test": "echo \"Error: no test specified\" && exit 1",
"start": "gulp"
},
The only other thing in my directory is a scripts folder with an app.js and an home.js file in it. Obviously once this task is run, the build folder is created (if it wasn't already there yet) and the two aforementioned files are copied into it.
You can see I'm only looking for scripts in the root directory's first level folder called scripts, so I shouldn't have an infinite loop by referencing changes on the same set of scripts. Also, even if I'm explicit, and point to exactly one particular file with a relative path such as ./scripts/home.js this still happens.
I'm anticipating being embarrassed, but I'm utterly confused.
A few things I've picked up on which could be causing some errors.
EDIT -
Try the watch plugin npm install --save-dev gulp-watch
// Try and declare your plugins like this for now.
var gulp = require('gulp'),
watch = require('gulp-watch');
// Provide a callback, cb
gulp.task('scripts', function(cb) {
// Dont use ./ on your src as watch can have a problem with this
return gulp.src('scripts/**/*.js')
.pipe(gulp.dest('./build/scripts'), cb); // call cb and dont forget ;
});
// remove ./ on watch
gulp.task('watch', function() {
gulp.watch('scripts/**/*.js', ['scripts']);
});
gulp.task('default', ['watch']);
So that is pretty weird behaviour but this should do the trick.
The only time I use ./ within gulp is on my dest.
Also just remember that gulpfile is just a JS so remember your semicolon, etc.
I cannot guarantee the resolution here, but I had a two variables that changed when this was resolved:
I upgraded my Parallels VM application (on an Apple PowerBook) from
version 10 -> 11.
I reinstalled Windows 10 using another license for a current version
(the previous one was a licensed dev or early release version).
My code, Node version, devDependencies and versions were identical.
I've installed nodejs and gulp to auto compile the bootstrap less file (v 3.3.5) with gulp-watch-less module.
Everything is working fine expect one thing: I have to stop and start gulp to regenerate bootstrap.css.
For information, Gulp is detecting that a .less file included in bootstrap.less is modified, I have the following message:
[23:14:40] Starting 'default'...
[23:14:42] Finished 'default' after 2.04 s
[23:16:42] LESS saw variable-overrides.less was changed
[23:16:42] LESS saw variable-overrides.less was changed
[23:16:42] LESS saw bootstrap.less was changed:by:import
[23:16:42] LESS saw bootstrap.less was changed:by:import
But when I open the bootstrap.css file i don’t see the changes until I stop and start gulp again.
Here is the content of my gulpfile.js:
var gulp = require('gulp');
var watchLess = require('gulp-watch-less');
var less = require('gulp-less');
gulp.task('default', function () {
return gulp.src('./../../../../drupal8/sandbox/felicity/themes/octogone/less/bootstrap.less')
.pipe(watchLess('./../../../../drupal8/sandbox/felicity/themes/octogone/less/bootstrap.less'))
.pipe(less())
.pipe(gulp.dest('./../../../../drupal8/sandbox/felicity/themes/octogone/css'));
});
This code is from gulp-watch-less page
Can some one explain me why the bootstrap.css is not auto-re-generated?
I've solved the problem by using gulp-watch-less2
Gulp-watch-less wasn't compatible with gulp 3.9
I have a gulp script that concatenate and minify my JavaScript.
With the gulp-html-replace plugin I can replace all my JS dependancies by the concatened file in my index.html.
I end up with a dev version (/dev/index.html), with all the single JS files included (easier for debugging) and a production version,with all JS concatened (/prod/index.html).
For now I have a config flag (in NodeJS) in a config.js file and I do the following :
res.render(((config.build === 'prod') ? './prod' : './dev') + 'myPage')
But I'm not really happy with this solution as it adds a lot of code and it's easy to forget to write this code.
Is there a better solution ?
Does this solution take place in Gulp
(by havign a gulp prod and a gulp dev for example)
Or does it take place in Node (by setting up a virtual directory for example)
I am new to this npm/gulp/node workflow and not sure of what belongs where
The way I like to do it is by maintaining two separate versions for index.html.
index-development.html for dev environment and index-production.html for production environment.
The index-development.html includes all the scripts and css (non minified and concatenated) and index-production.html as minified and concatenated scripts and css links.
I construct index.html from gulp script.
By default the index-development.html will be deployed.
If I specify parameter p to the gulp script, it will deploy index-production.html
No need to update the file path of the file to be served in your express router.
First do
npm install yargs
In gulp, I include
var argv = require('yargs').argv;
Check if parameter p (gulp -p) is passed to the gulp (p for production) with
var isProduction = argv.p;
and then,
if(isProduction){
taskSequence = ['combineControllers','combineServices','productionsIndex','startServer'];
} else{
taskSequence = ['developmentIndex','startServer'];
}
gulp.task('default', taskSequence);
gulp.task('startServer', function(){
exec('npm start', function (err, stdout, stderr) {
console.log(stdout);
console.log(stderr);
cb(err);
});
});
gulp.task('productionsIndex', function(done) {
return gulp.src('./www/index-productions.html')
.pipe(concat('index.html'))
.pipe(gulp.dest('./public/'));
});
gulp.task('developmentIndex', function(done) {
return gulp.src('./www/index-development.html')
.pipe(concat('index.html'))
.pipe(gulp.dest('./public/'));
});
This way, your index.html file will be constructed dynamically without having to change of the code in your express and you can serve it like
res.render('index');
if you want to user myPage.html everywhere, just replace index.html and index in the code above with myPage.html and myPage.
EDIT:
To start your application in development environment, simply run gulp
To start your application in production environment, simply run gulp -p
Simple!
in your app initialization process you can set the path of your views.
app.set('views', process.cwd() + ((config.build === 'prod') ? '/prod' : '/dev'));
Now you can call the render function like this:
res.render('myPage');
I've started working with Gulp and the problem I'm having is getting gulp-chug to work properly.
I've followed everything in the documentation, telling my gulpfile to watch all gulpfiles within certain directories, whereas it only watches one file.
This is the code I have used following the documentation...
var gulp = require('gulp');
var chug = require('gulp-chug');
gulp.task('default', function () {
gulp.src('**/task_runner/gulpfile.js')
.pipe(chug());
});
I even tried to see if it makes a difference if I put the filepath in an array...
...
gulp.src(
[ '**/task_runner/gulpfile.js' ]
)
...
I also tried this (and a version without the array in gulp.src())...
...
gulp.src(
[ 'Project_01/task_runner/gulpfile.js', 'Project_02/task_runner/gulpfile.js' ]
)
...
...and it still does the same thing.
My file structure looks like this,
*root*
node_modules
gulpfile.js
package.json
Project_01
css
scss
task_runner
Project_02
css
scss
task_runner
All the gulpfiles work when running them individually, but I want them all to run at the same time within one cmd window with gulp-chug.
This is what my cmd looks like, which is showing that it's only watching Project_02,
C:\Users\WaheedJ\Desktop\UniServer\www\Practice\gulp>gulp
[14:19:40] Using gulpfile ~\Desktop\UniServer\www\Practice\gulp\gulpfile.js
[14:19:40] Starting 'default'...
[14:19:40] Finished 'default' after 6.37 ms
[gulp-chug] File is a buffer. Need to write buffer to temp file...
[gulp-chug] Writing buffer to Project_02\task_runner\gulpfile.tmp.1411996780120.
js...
[gulp-chug] Spawning process C:\Users\WaheedJ\Desktop\UniServer\www\Practice\gul
p\Project_02\task_runner\node_modules\gulp\bin\gulp.js with args C:\Users\Waheed
J\Desktop\UniServer\www\Practice\gulp\Project_02\task_runner\node_modules\gulp\b
in\gulp.js --gulpfile gulpfile.tmp.1411996780120.js default from directory C:\Us
ers\WaheedJ\Desktop\UniServer\www\Practice\gulp\Project_02\task_runner...
[gulp-chug](Project_02\task_runner\gulpfile.tmp.1411996780120.js) [14:19:42] Usi
ng gulpfile ~\Desktop\UniServer\www\Practice\gulp\Project_02\task_runner\gulpfil
e.tmp.1411996780120.js
[gulp-chug](Project_02\task_runner\gulpfile.tmp.1411996780120.js) [14:19:42] Sta
rting 'watch'...
[gulp-chug](Project_02\task_runner\gulpfile.tmp.1411996780120.js) [14:19:43] Fin
ished 'watch' after 18 ms
[14:19:43] Starting 'default'...
[14:19:43] Finished 'default' after 7.13 µs
What can I do to fix this?
I have the same thing happening. For now i employed this workaround :
gulp.task('default', ['one-gulpfile', 'another-gulpfile'], function () {});
gulp.task('one-gulpfile', function () { return gulp.src('./project-one/gulpfile.js').pipe(chug()); });
gulp.task('another-gulpfile', function () { return gulp.src('./project-another/gulpfile.js').pipe(chug()); });
Basically an empty default task, with dependencies on hard coded tasks that each, run one gulp file.
Of course not dynamic, and needs maintenance, but I got it going which is what i needed most at this point in time. I hope to see chug mature a bit more.