本文介绍了咕噜咕噜,然后缩小任务的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
- 减少.css文件的格式
- Minify生成的css文件
这是我的gulp文件:
var gulp = require('gulp'),
watch = require(gulp-watch),
less = require('gulp-less'),
cssmin = require('gulp-cssmin'),
rename = require('gulp-rename');
gulp.task('watch-less',function(){
watch({glob:'./*.less'},function(files){// watch any changes关于咖啡文件
gulp.start('compile-less'); //运行编译任务
});
watch({
glob:[' ./*.css','!./*。min.css']
},函数(文件){
gulp.start('minify-css'); //运行编译任务
});
});
$ b $ gulp.task('compile-less',function(){
gulp.src('./*。less')//文件路径
.pipe ('('error',function(err){
console.log(err);
}))
.pipe(gulp.dest('./') );
});
$ b $ gulp.task('minify-css',function(){
gulp.src([
'./*.css',
'!。 /*.min.css'
))
.pipe(cssmin()。on('error',function(err){
console.log(err);
}))
.pipe({(后缀:'.min'}))
.pipe(gulp.dest('./'));
})
gulp.task('default',['watch-less']);
当我开始时,只有第一步完成。
请帮助我。
解决方案
没有必要经过一段时间,对我来说便利的解决方案是:
var gulp = require('gulp'),
less = require('gulp-less'),
cssmin = require('gulp-cssmin'),
plumber = require('gulp-plumber'),
rename = require('gulp-rename');
$ b $ gulp.task('watch',function(){
gulp.watch('./ styles / *。less',['less']);
} );
$ b $ gulp.task('less',function(){
gulp.src('./ styles / *。less')
.pipe(plumber())$
.pipe(less())
.pipe(gulp.dest('./ styles /'))
.pipe(cssmin())
.pipe(rename({
后缀:'.min'
}))
.pipe(gulp.dest('./ styles'))
});
gulp.task('default',['less','watch']);
I have to make 2 steps in gulp:
- Make a .css file form less
- Minify generated css files
This is my gulpfile:
var gulp = require('gulp'),
watch = require("gulp-watch"),
less = require('gulp-less'),
cssmin = require('gulp-cssmin'),
rename = require('gulp-rename');
gulp.task('watch-less', function () {
watch({glob: './*.less'}, function (files) { // watch any changes on coffee files
gulp.start('compile-less'); // run the compile task
});
watch({
glob: ['./*.css', '!./*.min.css']
}, function(files) {
gulp.start('minify-css'); // run the compile task
});
});
gulp.task('compile-less', function () {
gulp.src('./*.less') // path to your file
.pipe(less().on('error', function(err) {
console.log(err);
}))
.pipe(gulp.dest('./'));
});
gulp.task('minify-css', function() {
gulp.src([
'./*.css',
'!./*.min.css'
])
.pipe(cssmin().on('error', function(err) {
console.log(err);
}))
.pipe(rename({suffix: '.min'}))
.pipe(gulp.dest('./'));
})
gulp.task('default', ['watch-less']);
When i start it only first step is done.Help me please.
解决方案
There is no needing after time, convinient solution for me was:
var gulp = require('gulp'),
less = require('gulp-less'),
cssmin = require('gulp-cssmin'),
plumber = require('gulp-plumber'),
rename = require('gulp-rename');
gulp.task('watch', function () {
gulp.watch('./styles/*.less', ['less']);
});
gulp.task('less', function () {
gulp.src('./styles/*.less')
.pipe(plumber())
.pipe(less())
.pipe(gulp.dest('./styles/'))
.pipe(cssmin())
.pipe(rename({
suffix: '.min'
}))
.pipe(gulp.dest('./styles'))
});
gulp.task('default', ['less', 'watch']);
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