我希望编写我的gulpfile.js扫描style.scss文件的themes目录,其目的是读取style.scss文件并在同一目录中写入相应的style.css和.min文件.我遇到的问题是我找不到编写css文件的方法而不知道目录是什么...我不会.
这可能与gulp.dest()有关吗?
tl; dr:基本上......如何确定正在处理的*.scss文件的当前路径,以便我可以将*.css文件放在同一目录中
// GULP variable declarations
var gulp = require('gulp'),
gutil = require('gulp-util'),
sass = require('gulp-ruby-sass'),
prefix = require('gulp-autoprefixer'),
minifycss = require('gulp-minify-css'),
rename = require('gulp-rename'),
concat = require('gulp-concat'),
uglify = require('gulp-uglify');
// Paths array
var path = {
scss: [
'docroot/profile/theme/**/*/style.scss',
],
watch_scss: [
'docroot/profile/theme/**/*.scss',
],
};
// Process SASS functionality
gulp.task('process-scss', function() {
return gulp.src(path.scss)
.pipe(sass({
compass: true,
style: 'expanded',
}))
.pipe(prefix(['last 2 versions']))
.pipe(concat('style.css'))
.pipe(gulp.dest('./relative/dir')
.pipe(rename({suffix: '.min'}))
.pipe(minifycss())
.pipe(gulp.dest('./relative/dir')
.on('error', gutil.log);
});
// Setup the …Run Code Online (Sandbox Code Playgroud)