I have the problem, is went I run the task, everything is ok, but never give me the style.css result or output.
var gulp = require('gulp'),
concat = require('gulp-concat'),
uglify = require('gulp-uglify'),
sass = require('gulp-sass'),
compass = require('gulp-compass'),
neat = require ('node-neat').includePaths,
bourbon = require('node-bourbon');
// Bourbon Compile
gulp.task("compileBourbon", function(){
gulp.src('./src/sass/bourbon.scss')
.pipe(sass({
includePaths: require('node-bourbon').includePaths,
style: 'compressed',
quiet: true
}))
.pipe(gulp.dest('./builds/development/css'));
});
Bourbon is a mixin & function library so simply using it won't actually output any code, similar to how defining a function doesn't actually run the function. Neat is the same way, it only defines things that can be called but doesn't actually make any code by itself.
You'll want to #import "bourbon"; and #import "neat"; and then write css that uses the imported libraries like…
// mystyles.scss
#import "bourbon";
.my-class {
#include position(relative, 5em 2em null null);
color: blue;
}
As a side note, you probably don't want to be importing/using bourbon and compass at the same time. There is a bit of over lap and weird things can happen if you use them both.
Related
First of all I'd like you guys to be gentle. I haven't been coding much in recent year and since gulp update when then changed syntax to writing functions and exporting I somehow made it work then and left with no changes up to this point, no clue if they changed something else. I've been happy with what it is right now, but I have no idea how to make it work the other way.
So anyway I'm working on a project right now, where there will be many htmls, and each one will have quite different styles, but some will be common. I want to make a main.scss file with common styles for each html, but I want to make a separate scss with styles specific to each html. This way in the end I want to have a separate css file made from a specific scss with same name combined with main.scss, so that it won't have to download a single large file, but only styles I need.
Example:
main.scss
01.scss
02.scss
03.scss
will compile to:
01.css ( main.scss + 01.scss )
02.css ( main.scss + 02.scss )
03.css ( main.scss + 03.scss )
This is my gulpfile right now:
const gulp = require('gulp');
const sass = require('gulp-sass');
const browserSync = require('browser-sync').create();
function style() {
return gulp.src('./scss/**/*.scss')
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('./css'))
.pipe(browserSync.stream());
}
function watch() {
browserSync.init({
server: {
baseDir: './'
}
});
gulp.watch('./scss/**/*.scss', style);
gulp.watch('./*.html').on('change', browserSync.reload);
gulp.watch('./js/**/*.js').on('change', browserSync.reload);
}
exports.style = style;
exports.watch = watch;
If you have an idea how to do it in a better way I would really appreciate it.
I think you will have to import your main.scss into each of your other files and exclude main.scss from your gulp.src.
function style() {
return gulp.src(['./scss/**/*.scss', '!./scss/**/main.scss'])
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('./css'))
.pipe(browserSync.stream());
}
'!./scss/**/main.scss' this negates or excludes that file from being passes into this task - I assumed main.scss was in the same folder as your other scss files, if that is not the case you will have to modify the path.
Then #import main.scss into each of your 01.scss, 02.scss, etc. files:
#import "main.scss"; // also assumes in same folder
You can put this import statement anywhere in the file, if it is first any of main.scss styles will be overridden by conflicting styles in the rest of the 0x.scss file. If you put the import statement at the end, then main.scss styles will override any previous conflicting styles.
Note: you should really be using #use instead of #import and gulp-dart-sass instead of gulp-sass at this point. See sass #use rule.
// in your gulpfile.js
const sass = require('gulp-dart-sass'); // once installed
#use "main.scss"; // must be at top of each scss file, such as 01.scss
I'm trying to pull in the Sass bootstrap source (.SCSS), make some customisations (via another .SCSS) and spit out a CSS file.
I've been trying to use Gulp to do this is VS2019 using gulp-sass. I've followed many tutorials and have come up with the following gulpfile.js
var gulp = require('gulp');
var sass = require('gulp-sass');
sass.compiler = require('node-sass');
gulp.task('sass', function () {
return gulp.src('./Main.scss')
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('./css'));
});
In my Main.css I have the following:
$theme-colors: ( "primary": #fd7e14 );
#import "//lib/bootstrap-4.4.1/scss/bootstrap";
body {
color: #5CFF54;
background: rgb(92,18,18);
height: 400vh;
}
However, the file that is being generated contains the following:
#import "//lib/bootstrap-4.4.1/scss/bootstrap";
body {
color: #5CFF54;
background: #5c1212;
height: 400vh; }
I was expecting it to pull all of the individual styles into the produced CSS file, not just add the import.
Can anyone point me in the right direction?
After much trawling of the internet I have discovered that this is expected behaviour, and is down to the way I am referencing the source bootstrap.scss file.
In short, I am referencing it with web path, this has the effect of adding an import statement to the produced .css file. If I change the reference to a filesystem path such as this:
#import "../lib/bootstrap-4.4.1/scss/bootstrap";
it functions as I had hoped and the produced .css file includes all of the definitions from bootstrap.scss.
I'm on node version 11 (although I've tried 10 and 9), I'm using gulp version 3.9.1 and gulp-cli version 2.0.1, gulp-sass version 4.0.2, and node-sass version 4.11.2. Here is my very simple scss file:
#import "normalize-scss/sass/normalize";
h1 {
color: red;
}
Here is a dead simple gulpfile I am testing with:
var gulp = require('gulp'),
sass = require('gulp-sass'),
connect = require('gulp-connect');
gulp.task('sass', function(){
gulp.src('src/scss/style.scss')
.pipe(sass())
.pipe(gulp.dest('src/css'));
});
gulp.task('serve', ['sass'], function () {
connect.server();
});
gulp.task('default', ['serve']);
When I start the server with a bad import path, like say #import "normalize-scss/asdf/sass/normalize";, I get an error saying the patch can't be found. However, when I use the actual import path and start my task (like what's above), I get no errors, yet I only get this in my style.css file:
h1 {
color: red; }
I've been banging my head against this for an hour. Any clues?
It's my own fault; normalize.scss as it is found at https://github.com/appleboy/normalize.scss requires an #include normalize(); to start working. I guess I had spent so much time on gulp debugging that I lost track of the simple things.
I want compile different css files based on different sass variables.
Im trying to use Gulp to achieve this
gulp.task('var 1', function() {
gulp.src(['styles/vars/var1.scss','styles/client.scss'])
.pipe(sass().on('error', sass.logError))
.pipe(concat('clientWithVar1.css'))
.pipe(gulp.dest('./public/'))
});
It doesnt seem to work and im getting a Sass error that a variable doesnt exist.
How can i achieve this? Any help is appreciated
You can try this way:
You have this dir structure:
- tmp (folder)
- _var1.scss
- _var2.scss
- client.scss
In client.scss add: #import "./tmp/var.scss
and add gulp task below, so you just copy and rename scss file with variables that you need and then you can compile your main scss file
var rename = require('gulp-rename');
gulp.task('copy', function() {
return gulp.src('./styles/_var1.scss')
.pipe(rename('var.scss'))
.pipe(gulp.dest('./styles/tmp'));
})
Assuming you've got a directory structure something like this:
vars/var1.scss
vars/var2.scss
vars/var3.scss
vars/var4.scss
styles/1/client.scss
styles/2/client.scss
styles/3/client.scss
styles/4/client.scss
And in each client.scss file you have a corresponding #import statement (e.g. #import "../../vars/var1/.scss";), you should be able to generate all stylesheets with a single task:
gulp.task('sass', function() {
gulp.src('styles/**/client.scss')
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('./public/'))
});
I think this should end up with public/1/client.css, public/2/client.css etc, but you may have to tweak the gulp.dest line to suit.
I'm using gulp-ruby-sass to compile my js and sass.
I ran into this error first TypeError: Arguments to path.join must be strings
Found this answer and it was because I was using sourcemaps with gulp-sass and the answer recommended using gulp-ruby-sass instead.
Next I tried to compile all my SASS files using this syntax:
gulp.task('sass', function () {
return sass('public/_sources/sass/**/*.scss', { style: 'compressed' })
.pipe(sourcemaps.init())
.pipe(concat('bitage_public.css'))
.pipe(sourcemaps.write('./maps'))
.pipe(gulp.dest('public/_assets/css'))
.pipe(livereload());
});
Which produced this error:
gulp-ruby-sass stderr: Errno::ENOENT: No such file or directory - public/_sources/sass/**/*.scss
I then noticed in the answer I found the author wrote that globes ** aren't supported yet:
Also keep in mind, as of this writing when using gulp-ruby-sass 1.0.0-alpha, globs are not supported yet.
I did more digging and found a way to use an Array to specify the paths to my SASS files, so then I tried the following:
gulp.task('sass', function () {
return sass(['public/_sources/sass/*.scss',
'public/_sources/sass/layouts/*.scss',
'public/_sources/sass/modules/*.scss',
'public/_sources/sass/vendors/*.scss'], { style: 'compressed' })
// return sass('public/_sources/sass/**/*.scss', { style: 'compressed' })
.pipe(sourcemaps.init())
.pipe(concat('bitage_public.css'))
.pipe(sourcemaps.write('./maps'))
.pipe(gulp.dest('public/_assets/css'))
.pipe(livereload());
});
But still I'm getting Errno::ENOENT: No such file or directory and it lists all the dirs I put into that array.
How do you compile SASS in multiple directories with gulp?
SASS source folder structure:
_sources
layouts
...scss
modules
...scss
vendors
...scss
main.scss
Figured it out!
Well not 100%, still not sure why the multiple path array didn't work.
Anyways so I forgot that in my main web.scss file I already had multiple import statements setup:
#import "vendors/normalize"; // Normalize stylesheet
#import "modules/reset"; // Reset stylesheet
#import "modules/base"; // Load base files
#import "modules/defaults"; // Defaults
#import "modules/inputs"; // Inputs & Selects
#import "modules/buttons"; // Buttons
#import "modules/layout"; // Load Layouts
#import "modules/svg"; // Load SVG
#import "modules/queries"; // Media Queries
So I didn't actually need to try use Gulp the way I was trying, I just needed to target that 1 .scss file directly. So I did that here:
// Compile public SASS
gulp.task('sass', function () {
return sass('public/_sources/sass/bitage_web.scss', { style: 'compressed' })
.pipe(sourcemaps.init())
.pipe(sourcemaps.write('./maps'))
.pipe(gulp.dest('public/_assets/css'))
.pipe(livereload());
});
Now it works because it sees a specific file to target and compile
I was having trouble using '*.scss' too
In the git documentation (https://github.com/sindresorhus/gulp-ruby-sass) they use this sintax:
gulp.task('sass', function(){
return sass('public/_sources/sass/',
{ style: 'compressed'})
.pipe(sourcemaps.init())
});
I tested it and it works, it compiles all the files within the folder.
Just in case someone has the same problem