I'm tryng to create a new gulp task to run into my application and look for all '.fragment.sass' files.
I wrote:
gulp.task('sassFragments', () => {
return gulp
.src('./src/**/*.fragment.sass')
.pipe(debug())
.pipe(sassGlob())
.pipe(sass({ outputStyle: 'expanded' })).on('error', sass.logError)
.pipe(concat('fragments_style.css'))
.pipe(gulp.dest('./build/assets/css'))
.pipe(browserSync.reload({ stream: true }));
})
but no fragments_style.css is created in /build/assets/css folder.
I have another task which does similar using src('./src/**/*.sass') to generate a style.css file and works great!
I think there is a issue with .src method, that is not matching this '.fragment.sass' pattern.
Can anyone help me?
Gulp version: 3.9.1
I have a shared SCSS source files which must be compiled and copied into different project folders.
I have a build task which calls 2 tasks, clean and styles(to compile/minify and copy to build folder).
My source SCSS files are shared between all websites however the destination folders are different.
I would like to be able to run: build websiteA and then clean build folder inside websiteA and compile files from a shared folder and copied to build folder inside Website A.
var assetsDir = '_Assets';
var buildStyleWebsiteA = 'WebsiteA/Assets/build';
var buildStyleWebsiteB = 'WebsiteB/Assets/build';
gulp.task('clean-websiteA', function (cb) {
return del([buildStyleWebsiteA ], cb);
});
gulp.task('styles-websiteA', ['clean-websiteA'], function () {
return gulp.src(assetsDir + '/**/*.scss')
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(autoprefixer())
.pipe(gulp.dest(buildStyleWebsiteA + '/css'))
.pipe(concat('styles.css'))
.pipe(cleanCss())
.pipe(sourcemaps.write())
.pipe(rename({ suffix: '.min' }))
.pipe(gulp.dest(buildStyleWebsiteA + '/min/'))
.pipe(liveReload());
});
gulp.task('build-websiteA', ['styles']);
PS: I also have same tasks for websiteB (build-websiteB, clean-websiteB, and style-websiteB).
So I ended up with repetitive code and I know there must be a better way.
What I would like to have is provide website name as a parameter for gulp command and then it runs clean and style using correct folder related to that website.
How can I refactor my code to achieve that?
Thanks
I would use environment variables to accomplish this rather than arguments.
var buildDir = process.env.BUILD_DIR + '/Assets/build';
Then you would call it as:
BUILD_DIR=websiteA gulp build
I am using Gulp to compile my bootstrap.less file into a bootstrap.css file. All the bootstrap LESS files are in a "less" folder and all of the CSS files are in a CSS folder. All of the bootstrap less files are #import'd into bootstrap.less and then that file is compiled to bootstrap.css. However I have one custom.less file that is in the bootstrap LESS folder. I would like to compile into a custom.css file in the CSS folder. I cannot get Gulp to do this. Here is my code.
var gulp = require('gulp');
var less = require('gulp-less');
var browserSync = require('browser-sync').create();
gulp.task('less', function() {
return gulp.src(['./less/bootstrap.less', './less/custom.less'])
.pipe(less())
.pipe(gulp.dest("./css"))
.pipe(browserSync.reload({stream: true}));
});
gulp.task('serve', function(){
browserSync.init({
server: {
baseDir: './'
}
});
gulp.watch('./less/*.less', ['less']);
gulp.watch(['./**/*.html', './js/custom.js']).on('change',
browserSync.reload);
});
gulp.task( 'default', ['less', 'serve']);
What I usually do is compile all .less or .scss files into css folder. You will end up with a few empty files (a good example would be .less files that contain variables) but on the other hand you can create a gulp task to clear those files out. I use the following code to compile:
return gulp.src("Styles/scss/*.scss")
.pipe(sourcemaps.init())
.pipe(sass())
.pipe(sourcemaps.write('.'))
.pipe(gulp.dest("Styles/css"))
.pipe(reload({stream: true}));
I'm also generating sourcemaps here but you can remove that. If you would like to discover more checkout this post here
Hope this helps ;)
I've been trying to get gulp sass and gulp sourcemaps to do exactly what I want and I'm finding it hard. I want to take a sass entry file (src/sass/index.scss), generate an output file (dist/css/index.css) and a separate sourcemap for that index file (dist/css/index.css.map) which has a sourceRoot set to the project base (absolute path: /home/damon/projects/test) and the sourcemap entries to be relative to that path.
Here's what I tried:
attempt 1: straight example code from gulp-sass:
var sassEntry = 'src/sass/index.scss';
gulp.src(sassEntry)
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(sourcemaps.write())
.pipe(gulp.dest('dist/css'));
Outcome: this inlines the sourcemap into the CSS file so I can't tell if it's right or not.
attempt 2: write it to separate file
gulp.src(sassEntry)
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(sourcemaps.write('.'))
.pipe(gulp.dest('dist/css'));
Outcome: writes a separate sourcemap, but the sourceRoot says '/sources/' (WTF is that?!, it doesn't exist and I never configured it)
and the paths are all relative to the sass entry file, not the project base, which is also going to be meaningless when my browser tries to locate the source files.
attempt 3: try to fix the sourceroot (also I found includeContent: false)
gulp.src(sassEntry)
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(sourcemaps.write('.',{includeContent: false, sourceRoot: __dirname}))
.pipe(gulp.dest('dist/css'));
Outcome: the sourceroot is now my working folder which is nice, the content isn't included which is nice, but the files in the sourcemap are still relative to the sass entry file not to the sourceRoot, so my map is still useless
attempt 4: Set the gulp.src base
gulp.src(sassEntry, { base: __dirname })
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(sourcemaps.write('.',{includeContent: false, sourceRoot: __dirname}))
.pipe(gulp.dest('dist/css'));
Outcome: Infuriatingly, setting the base on gulp.src fixes the sourcemap - sourceRoot is still correct and the source file paths are relative to the sourceRoot, BUT it now outputs to dist/css/src/sass/index.css which is wrong. WTF!
attempt 5: use absolute paths
gulp.src(sassEntry, { base: __dirname })
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(sourcemaps.write('.',{includeContent: false, sourceRoot: __dirname}))
.pipe(gulp.dest(__dirname + '/dist/css'));
Outcome: no change, it still outputs to the same deep structure in dist.
If anyone can enlighten me on how to do this I would be forever grateful.
While Sven's answer is perfectly good, I also found an answer to my own question by getting a deeper understanding of how gulp works (which I was trying to avoid), and apparently gulp stores each matched file with a path, so adding:
{ base: __dirname }
in the gulp.src makes it that each matched file has the full path from the base, which then causes them to output with the full relative path from wherever you set the base to. The solution I ended up with was to use gulp-flatten, which removes those relative paths from files in the pipeline, so my eventual function looked like this:
gulp.src(sassEntry, { base: __dirname })
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(sourcemaps.write('.',{includeContent: false, sourceRoot: __dirname}))
.pipe(flatten())
.pipe(gulp.dest(__dirname + '/dist/css'));
easy once you understand more about what it's trying to do I guess.
Since your attempt 4 does everything you want except place the resulting files in the wrong location, the easiest fix would be to just change that location with gulp-rename after the sourcemaps have been generated:
gulp.src(sassEntry, { base: __dirname })
.pipe(sourcemaps.init())
.pipe(sass().on('error', sass.logError))
.pipe(sourcemaps.write('.', {includeContent: false, sourceRoot: __dirname}))
.pipe(rename({dirname:''}))
.pipe(gulp.dest('dist/css'));
I'm trying to write a simple gulp task that takes a scss file called manifest.scss and after compiling and minifying the file it saves the result into a destination folder as app.css
The following task does almost everything I want beside renaming the file (the output is build/css/manifest.css)
gulp.task('sass', function() {
gulp.src("src/sass/manifest.scss")
.pipe(sass({ style: 'compressed' }))
.pipe(minifyCSS())
.pipe(gulp.dest('build/css'));
});
So, I have tried gulp-rename and I have update the task as follows:
gulp.task('sass', function() {
gulp.src("src/sass/manifest.scss")
.pipe(sass({ style: 'compressed' }))
.pipe(minifyCSS())
.pipe(rename('app.css'))
.pipe(gulp.dest('build/css'));
});
This produces the build/css/app.css file but it is totally blank.
How can I rename the compiled file?
Thanks