I am having an issue with my gulp build file with the sass imports.
I am using gulp-importify to create a build.scss file that has the proper order of my dependencies, third party css, and my custom css.
The Importify task
module.exports = function() {
APP.gulp.src(APP.concat.sass)
.pipe(APP.tasks.importify('build.scss', {
cssPreproc: 'scss'
}))
.pipe(APP.gulp.dest('build/sass/'));
};
My folder structure is as follows:
gulp
- bower_components
- build
- sass
- build.scss
- modules
-forms.scss
The build.scss file which contains the imports has the following paths
#import "_bootstrap-sprockets";
#import "variables.scss";
#import "modules/form.scss";
but the bower components are back two directories.
#import "../../bower-components/bootstrap-sass/assets/stylesheets/_bootstrap-sprockets";
#import "variables.scss";
#import "modules/form.scss";
is there any other way or method to setting the full relative/correct import path in Gulp?
I found the solution. I added a base path through gulp.src as follows:
module.exports = function() {
APP.gulp.src(APP.concat.sass,
{base: 'build/sass'})
.pipe(APP.tasks.imports('build/sass/build.scss', {
cssPreproc: 'scss'
}))
.pipe(APP.gulp.dest('.'));
};
You can also use the global process object available in Node to get the current working directory. So you can set the base to:
{base: process.cwd() }
Related
I have a scss folder I'm watching. I'm including all partials into main.scss and I'd like the generated file to be named style.css
Below is a portion of my gulpfile.js
const rename = require("gulp-rename");
const files = {
scssPath: './src/scss/**/*.scss',
jsPath: 'src/js/**/*.js'
}
function scssTask() {
return src(files.scssPath)
.pipe(sass())
.pipe(rename('style.css'))
.pipe(postcss([ autoprefixer(), cssnano()]))
.pipe(dest('web/assets/css'));
}
// default
exports.default = series(
parallel(scssTask, jsTask),
watchTask
);
The rename part doesn't seem to have any affect. The file gets generated into main.css
Sure, I could just rename my main.scss to style.scss, but I like having "main" for the sass files and "style" for the generated file the browser will use
Also, here's my main.scss file:
#import 'utilities/variables';
#import 'base/normalize';
#import 'base/typography';
I have two files:
1 - common.styl (this file has imports that will be used on all pages of my project.)
#import 'utils/variables.styl'
#import 'utils/fonts.styl'
#import 'utils/mixin.styl'
2 - home.styl (this file is only for home section of my project and depends on common.styl)
body
font-family CoolDown
.box
background $commonBg
In my gulpfile I've created two tasks, one to compile the common.styl and another to compile all the pages of my project.
Common task:
gulp.task('commonCSS', function () {
return gulp.src('src/styles/common.styl')
.pipe($.plumber())
.pipe($.stylus({
'include css': true,
use: [jeet(), nib(), rupture()],
import: ['jeet', 'nib', 'rupture']
}))
.pipe($.autoprefixer(stylBrowsers))
.pipe(gulp.dest(paths.build + 'css'))
.pipe(reload({ stream: true }));
});
My problem is the pages task, this only works if I put the common.styl file in the src path and concat them. But doing this way I need to put a name in the file. I would like to use the name of the current .styl being compiled.
gulp.task('pagesCSS', ['commonCSS'], function () {
return gulp.src(['src/styles/common.styl', 'src/styles/pages/**/*.styl'])
.pipe($.plumber())
.pipe($.concat('**page.css**')
.pipe($.stylus({
'include css': true,
use: [ jeet(), nib(), rupture() ],
import: ['jeet', 'nib', 'rupture']
}))
.pipe($.autoprefixer(stylBrowsers))
.pipe($.minifyCss())
.pipe(gulp.dest(paths.build + 'css'))
.pipe(reload({ stream: true }));
});
The question is: There is a way to include the common.styl to be used by pagesCSS task?
Maybe I'm missing something or using the wrong solution here.
Can't you just #import or #require common.styl in those files? #import "../../common" and Stylus will include them for every page, no need for concat.
Or...
You can use import option you're already have in your config object. Stylus will include common.styl at the beginning of each file by itself.
First add paths: ['node_modules', 'src/styles']. This way Stylus will know how to resolve paths for imports. You can probably skip this step if you provide full path in next one.
Now you can add your common.styl to import: ['jeet', 'nib', 'rupture', 'common'].
I'm using this configuration for my variables so I don't have to include them in each file.
Full example should look somewhat like that:
gulp.task('pagesCSS', ['commonCSS'], function () {
return gulp.src('src/styles/pages/**/*.styl') // <-- only pages styles are piped
.pipe($.plumber())
// .pipe($.concat('**page.css**') // <-- remove this line
.pipe($.stylus({
'include css': true,
use: [ jeet(), nib(), rupture() ],
paths: ['node_modules', 'src/styles'] // <-- resolve your styles path
import: ['jeet', 'nib', 'rupture', 'common'] // <-- add your shared file
}))
.pipe($.autoprefixer(stylBrowsers))
.pipe($.minifyCss())
.pipe(gulp.dest(paths.build + 'css'))
.pipe(reload({ stream: true }));
});
I would like to create a resources.zip file which will contain css/styles.css.
So far I have got most of this working, the only problem is the archive only contains the styles.css file and not its parent directory css.
gulpfile.js
const gulp = require('gulp');
const zip = require('gulp-zip');
gulp.task('default', () => {
return gulp.src('css/*')
.pipe(zip('resources.zip'))
.pipe(gulp.dest('build'));
});
I think you need to setup the base for the gulp.src:
gulp.src('css/*', {base: '.'})
This is because the default base is:
Default: everything before a glob starts (see glob2base)
source. Zipped file path: zip.
I was learning gulp. I had come thru following code.
wrench.readdirSyncRecursive('./gulp').filter(function(file) {
return (/\.(js|coffee)$/i).test(file);
}).map(function(file) {
require('./gulp/' + file);
});
Can somebody help me to understand what above code is useful for?
This will load all js or coffee files in the gulp directory in order to load all gulp tasks, so you dont have to manually import new gulp tasks, just create whatevergulptask.js inside '/gulp' directory and you can use it from the command line.
Another advange of doing it like this is that you dont have a huge gulpfile.js with millions of tasks and lines of code, as instead you have a whatevergulptask.js per TASK, just a good practice cos the gulpfile grows pretty fast
Example gulpfile.js
/**
* Welcome to your gulpfile!
* The gulp tasks are splitted in several files in the gulp directory
* because putting all here was really too long
*/
'use strict';
var gulp = require('gulp');
var wrench = require('wrench');
/**
* This will load all js or coffee files in the gulp directory
* in order to load all gulp tasks
*/
wrench.readdirSyncRecursive('./gulp').filter(function (file) {
return (/\.(js|coffee)$/i).test(file);
}).map(function (file) {
require('./gulp/' + file);
});
/**
* Default task clean temporaries directories and launch the
* main optimization build task
*/
gulp.task('default', ['clean'], function () {
gulp.start('build');
});
YOUR FOLDER STRUCTURE
gulp/
build.js
whatevergulptask.js
...
I'm using gulp-durandal to build our durandal app. It fails on our first module which has a depeendecy to knockout through:
define(['knockout',....
[09:35:27] Durandal Error: ENOENT, no such file or directory 'C:\xxxxx\app\knockout.js'
In module tree:
company/viewmodels/edit
at Object.fs.openSync (fs.js:438:18)
I have knockout defined as a patch in config.js (standard requirejs way) but it seems gulp-durandal does not resolve paths from config.js ?
'knockout': '../Scripts/lib/knockout/knockout-2.3.0',
How do you get gulp-durandal to use our config paths instead of trying to resolve the modules directly under the app folder ? I tried using options.extraModules but that only allows you to add paths to modules, not symbolic names for the module so that doesn't seem to be the correct way.
The basic structure of my durandaljs app follows the standard guidelines I believe, I have a config.js and main.js under the App folder.
My config.js:
define([], function() {
return {
paths: {
'text': '../Scripts/lib/require/text',
'durandal': '../Scripts/durandal',
'plugins': '../Scripts/durandal/plugins',
My main.js
require(['config'], function(config) {
require.config(config);
require(['durandal/system', 'durandal/app', 'durandal/viewLocator', 'plugins/widget', 'custombindings'],
function(system, app, viewLocator, widget) {
..... app code here.....
}
gulpfile.js:
var gulp = require('gulp');
var durandal = require('gulp-durandal');
require(['App/config'], function(config){
console.log('loaded config');
});
gulp.task('durandal', function(){
durandal({
baseDir: 'app', //same as default, so not really required.
main: 'main.js', //same as default, so not really required.
output: 'main.js', //same as default, so not really required.
almond: true,
minify: true,
require:true
})
.pipe(gulp.dest('dir/to/save/the/output'));
});
I guess the question is how do I load my config.js paths into gulp so the paths are resolved correctly ? I tried:
var gulp = require('gulp');
var durandal = require('gulp-durandal');
require(['App/config'], function(config){
console.log('loaded config');
});
But it seems require only wants a string as input (I guess require function in gulp != require from require.js)
I believe the issue is that your gulp-durandal task needs configuration to mimic the config.js file. If you need further assistance please provide more code from your gulp-durandal task.