I have this project in Symfony
src
ProjectName
AdminBundle
Resources
public
css
style1.css
style2.ccs
sass
style1.scss
style2.scss
AdvertiserBundler
..
CommonBundle
..
PublicBundle
..
WebmaserBundle
..
All bundles have the same structure. All *.scss
files are in Resources/sass/
and compiled *.css
files are in Resources/public/css/
. How I can put all files from A to B with one function? I try work with base
, but nothing happens.
This is gulpfile.js
var gulp = require('gulp'),
sass = require('gulp-sass'),
autoprefixer = require('gulp-autoprefixer'),
minifycss = require('gulp-minify-css')
;
gulp.task('styles', function() {
gulp.src('src/ProjectName/*Bundle/Resources/sass/*.scss', {
base: 'sass'
})
.pipe(sass({
outputStyle: 'compressed',
errLogToConsole: true
}))
.pipe(autoprefixer('last 5 version', 'ie 9'))
.pipe(minifycss())
.pipe(gulp.dest('public/css/'))
;
});
gulp.task('default', ['styles'], function() {
gulp.watch('src/ProjectName/*Bundle/Resources/sass/*.scss', ['styles']);
});
I don't want write like this five times:
....
gulp.src('src/ProjectName/AdminBundle/Resources/sass/*.scss')
.pipe(gulp.dest('src/ProjectName/AdminBundle/Resources/public/css'))
....