I am very new to gulp and sass, but trying to create a gulpfile that will take and scss file and turn it into a css file (this seems like it is a very common thing that lots of people do). I am following along this tutorial - https://youtu.be/nusgoj74a3Y?t=1301 - but it seems to be a bit outdated.
My directory looks like this
-xxx
--gulpfile
--src
---Assets
----scss
-----default.scss
----css
My gulpfile looks like this:
'use strict';
//dependencies
var gulp = require('gulp');
var sass = require('gulp-sass');
var minifyCSS = require('gulp-clean-css');
var uglify = require('gulp-uglify');
var rename = require('gulp-rename');
var changed = require('gulp-changed');
//////////////////
// - SCSS/CSS - //
//////////////////
var SCSS_SRC = './src/Assets/scss/**/*.scss';
var SCSS_DEST = './src/Assets/css';
//compile css
gulp.task('compile_scss', function(){
gulp.src(SCSS_SRC)
.pipe(sass().on('error', sass.logError))
.pipe(minifyCSS())
.pipe(rename({ suffix: '.min' }))
.pipe(changed(SCSS_DEST))
.pipe(gulp.dest(SCSS_DEST));
});
//detect changes in SCSS
gulp.task('watch_scss', function() {
return gulp.watch(SCSS_SRC, gulp.series('compile_scss'));
});
//run tasks
gulp.task('default', gulp.series('watch_scss'));
and when I run it I get this:
[20:39:12] Using gulpfile ~/xxx/xxx/xxx/xxx/gulpfile.js
[20:39:12] Starting 'default'...
[20:39:12] Starting 'watch_scss'...
I believe it is supposed to finish these tasks not just start them.
Also I believe it is supposed to take the scss file from the scss directory and then put a css file in the css directory which it does not do.
Any help would be great. Thanks.
you are triggering a watch there, try modifying any of your scss files, and watch the terminal transpile to CSS ;)
you could add alternatively something like
gulp.task('getCSS', gulp.series('compile_scss'));
and then run gulp getCSS to get the CSS
Related
I am using GULP to compile my scss files and trying trying to compile it down into a single styles.css file. However when I run gulp sass it converts the scss files down to css but also replicates them instead of compiling them all down into one single file.
gulpfile.js
var gulp = require('gulp');
var sass = require('gulp-sass');
var sourcemaps = require('gulp-sourcemaps');
gulp.task('sass', function () {
return gulp.src('./src/sass/**/*.scss')
.pipe(sourcemaps.init())
.pipe(sass.sync({outputStyle: 'compressed'}).on('error', sass.logError))
.pipe(sourcemaps.write())
.pipe(gulp.dest('./src/css'));
});
gulp.task('sass:watch', function () {
gulp.watch('./src/sass/**/*.scss', ['sass']);
});
file structure
--gulpfile.js
--src
--css
--sass
--themes
--module.scss
--tm-1.scss
--styles.scss
when i run gulp sass the css output ends up being:
--css
--styles.css
--themes
--module.css
--tm-1.css
which all I want here is one single styles.css file and cant seem to understand what I'm missing here.
I have fixed the issue by changing the line
return gulp.src('./src/sass/**/*.scss')
to:
return gulp.src('./src/sass/styles.scss')
Using gulp-sass automatically calls the file that is compiled to css style.css. Does anyone know how this filename can be modified? i.e. so I can change it to style2.css
Please see gulpfile.js below.
var gulp = require('gulp'),
sass = require('gulp-sass');
gulp.task('sass', function () {
return gulp
.src('./dev/style.scss')
.pipe(sass())
.pipe(gulp.dest('./prod/css'));
});
gulp.task('default', ['sass']);
Try gulp-rename - looks pretty easy.
I just started to use Gulp. This code is used to compress a scss file to css. I can't figure out how to keep one of the css files uncompressed.
This is what I want it to be:
/assets/css/custom.css
/assets/css/custom.min.css
Code:
var gulp = require('gulp');
var sass = require('gulp-sass');
gulp.task('styles', function() {
gulp.src('assets/scss/custom.scss')
.pipe(sass({outputStyle: 'compressed'}))
.pipe(gulp.dest('./assets/css/'));
});
gulp.task('default',function() {
gulp.watch('assets/scss/**/*.scss',['styles']);
});
What you could do is expand your task so that it writes out the uncompressed version, then compresses it, renames it, then finally writes it out again.
var gulp = require('gulp');
var sass = require('gulp-sass');
var rename = require('gulp-rename');
gulp.task('styles', function() {
gulp.src('assets/scss/custom.scss')
.pipe(sass())
.pipe(gulp.dest('./assets/css/'))
.pipe(sass({outputStyle: 'compressed'}))
.pipe(rename({ suffix: '.min' }))
.pipe(gulp.dest('./assets/css/'));
});
Note that you'll need to add the gulp-rename plugin to your dependencies. You could also use a dedicated minifier (such as gulp-minify-css) rather than the SASS plugin to minify your CSS.
I have the need to compile a SASS file to a CSS file when saved, without having to compile every SASS file to a single CSS file.
I need the ability to:
- Run a 'watch' on a directory
- If a file is saved, a CSS of it's name is created. Example: 'main.scss' compiles to 'main.css'.
- It should not compile every single SASS if it doesn't need to.
The goal is to optimize the development process to avoid compiling every single SASS file in a directory when 'watching'.
My current SASS task looks a bit like this and results in a single CSS file:
//Compile Sass
gulp.task('styles', function() {
return gulp.src('app/scss/styles.scss')
.pipe(plugins.sass({ includePaths : [paths.sass], style: 'compressed'})
.pipe(plugins.autoprefixer('last 2 version'))
.pipe(plugins.rename({suffix: '.min'}))
.pipe(plugins.minifyCss())
.pipe(gulp.dest('build/css'));
});
Looks like gulp-changed is what you're looking for:
https://github.com/sindresorhus/gulp-changed
You add it as a dependency with npm install --save-dev gulp-changed and plug it into your gulpfile. From the gulp-changed ReadMe:
var gulp = require('gulp');
var changed = require('gulp-changed');
var ngAnnotate = require('gulp-ng-annotate'); // just as an example
var SRC = 'src/*.js';
var DEST = 'dist';
gulp.task('default', function () {
return gulp.src(SRC)
.pipe(changed(DEST))
// ngAnnotate will only get the files that
// changed since the last time it was run
.pipe(ngAnnotate())
.pipe(gulp.dest(DEST));
});
I have a bunch of Stylus files in './styles/stylus/**.styl' and a bunch of CSS files in './styles/css/**.css'.
How do I use Gulp to compile the Stylus files, concat the result with all of the CSS files and output it to './styles/out.css'?
You can use gulp-filter like:
var gulp = require('gulp');
var stylus = require('gulp-stylus');
var concat = require('gulp-concat');
var Filter = require('gulp-filter');
gulp.task('css', function () {
var filter = Filter('**/*.styl', { restore: true });
return gulp.src([
'./styles/stylus/**.styl',
'./styles/css/**.css'
])
.pipe(filter)
.pipe(stylus())
.pipe(filter.restore)
.pipe(concat('out.css'))
.pipe(gulp.dest('./styles'));
});
You should be able to create two separate streams of files and merge them with event-stream:
var es = require('event-stream');
gulp.task('styles', function() {
var stylusStream = gulp.src('./styles/stylus/**.styl')
.pipe(stylus());
return es.merge(stylusStream, gulp.src('./styles/css/**.css'))
.pipe(concat('out.css'))
.pipe(gulp.dest('./styles'));
});
Within your Stylus files, you can just #require CSS files:
#require 'something.css'
Then there’s no need for concatenation or any other complexity in Gulp, and your file load order is set explicitly within the .styl files.
I just found a great solution to this: use #require for each of the other .styl files in your main.styl file, and just watch gulp.src('src/css/main.styl')