让gulp动态地在与处理后的scss文件相同的目录中写入css [英] Have gulp dynamically write css in the same directory as the processed scss file
问题描述
我正在寻找写我的gulpfile.js扫描style.scss文件的主题目录,其想法是读取style.scss文件并写入相应的style.css& .min文件放在同一个目录下。我遇到的问题是,我无法找到一种方法来编写css文件,而不知道目录是什么......我不会。
这是可能的与gulp.dest()?
tl; dr:本质上......我如何确定正在处理的* .scss文件的当前路径以便我可以将* .css文件放在同一个目录中
gulpfile.js
gutil = require('gulp-util'),$ b $ ('gulp-rubify-css'),
rename = require('gulp-rename'),
concat = require('gulp-concat'),
uglify = require('gulp-uglify');
//路径数组
var path = {
scss:[
'docroot / profile / theme / ** / * / style.scss',
$,
watch_scss:[
'docroot / profile / theme / ** / *。scss',
],
};
//处理SASS功能
gulp.task('process-scss',function(){
return gulp.src(path.scss)
.pipe (前缀(['后2个版本')))$ b $(bss({
指南针:true,
style:'expanded',
}))
.pipe b.pipe(concat('style.css'))
.pipe(gulp.dest('./ relative / dir')
.pipe(rename({suffix:'.min'}) )
.pipe(minifycss())
.pipe(gulp.dest('./ relative / dir')
.on('error',gutil.log);
)};
//设置gulp WATCH功能
gulp.task('default',function(){
gulp.start('process-scss');
gulp.watch(path.watch_scss,['process-scss']);
});
只需添加,就可以使用gulp.dest()以下面的代码片段提供文件的基础URL。
$ b
https:// gi thub.com/gulpjs/gulp/blob/master/docs/API.md#path
gulp.dest(函数(文件){
返回file.base;
}
编辑: :
$ b
//文件路径变量声明
var stylePath =;
var scriptPath =;
//路径数组
var path = {
scss:[
'docroot / profile / theme / ** / style / scss / style.scss',
],
watch_scss:[
'docroot / profile / theme / ** / style / scss / ** / *。scss',
],
theme_base :[
'docroot / profile / theme /',
],
};
//处理SASS功能
gulp.task('process-scss',function(){
return gulp.src(path.scss)
.pipe (gulp.dest(function(file){
var relative = file.relative.split(/);
stylePath = path.theme_base + relative [0] +'/ style /';
返回file.base;
)))
.pipe(sass({
compass:true,
style:'expanded',
}))
.pipe(prefix(['last 2 versions']))
.pipe(concat('style.css'))
.pipe(gulp.dest(function(){return stylePath;}))
.pipe(rename({suffix:'.min'}))
.pipe(minifycss())
.pipe(gulp.dest(function(){ return stylePath;}))
.on('error',gutil.log);
});
I am looking to write my gulpfile.js scan the themes directory for style.scss files, and the idea is to read the style.scss file and write the corresponding style.css & .min file in the same directory. The issue that i'm having is that I can't find a way to write the css file without knowing exactly what the directory is... which I will not.
Is this possible with the gulp.dest()?
tl;dr: Essentially... how can I determine the current path of the *.scss file being processed so that I can place the *.css file in the same directory
gulpfile.js
// 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 gulp WATCH functionality
gulp.task('default', function() {
gulp.start('process-scss');
gulp.watch(path.watch_scss, ['process-scss']);
});
Just to add, it turns out that you can use gulp.dest() to provide the file's base url with the following snippet.
https://github.com/gulpjs/gulp/blob/master/docs/API.md#path
gulp.dest(function(file) {
return file.base;
}
edit: My final gulp task looked like this:
// File path variable declarations
var stylePath = "";
var scriptPath = "";
// Paths array
var path = {
scss: [
'docroot/profile/theme/**/style/scss/style.scss',
],
watch_scss: [
'docroot/profile/theme/**/style/scss/**/*.scss',
],
theme_base: [
'docroot/profile/theme/',
],
};
// Process SASS functionality
gulp.task('process-scss', function() {
return gulp.src(path.scss)
.pipe(gulp.dest(function(file) {
var relative = file.relative.split("/");
stylePath = path.theme_base + relative[0] + '/style/';
return file.base;
}))
.pipe(sass({
compass: true,
style: 'expanded',
}))
.pipe(prefix(['last 2 versions']))
.pipe(concat('style.css'))
.pipe(gulp.dest(function() { return stylePath; }))
.pipe(rename({suffix: '.min'}))
.pipe(minifycss())
.pipe(gulp.dest(function() { return stylePath; }))
.on('error', gutil.log);
});
这篇关于让gulp动态地在与处理后的scss文件相同的目录中写入css的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!