問題描述
我有一個(gè)類似的變量
var files = {
'foo.css': 'foo.min.css',
'bar.css': 'bar.min.css',
};
我希望 gulp 為我做的是 minify 文件，然后為我 rename.
What I want the gulp to do for me is to minify the files and then rename for me.
但任務(wù)當(dāng)前寫為(針對(duì)一個(gè)文件)
But the tasks is currently written as (for one file)
gulp.task('minify', function () {
gulp.src('foo.css')
.pipe(minify({keepBreaks: true}))
.pipe(concat('foo.min.css'))
.pipe(gulp.dest('./'))
});
如何重寫以使其與上面定義的變量 files 一起使用?
How to rewrite so it work with my variable files defined above?
推薦答案
您應(yīng)該能夠使用 Glob 為您的 src 選擇所需的任何文件，而不是在對(duì)象中定義它們，這應(yīng)該會(huì)簡化您的任務(wù).此外，如果您希望將 css 文件縮小為單獨(dú)的文件，則無需將它們連接起來.
You should be able to select any files you need for your src with a Glob rather than defining them in an object, which should simplify your task. Also, if you want the css files minified into separate files you shouldn't need to concat them.
var gulp = require('gulp');
var minify = require('gulp-minify-css');
var rename = require('gulp-rename');
gulp.task('minify', function () {
gulp.src('./*.css')
.pipe(minify({keepBreaks: true}))
.pipe(rename({
suffix: '.min'
}))
.pipe(gulp.dest('./'))
;
});
gulp.task('default', ['minify'], function() {
});
這篇關(guān)于CSS 使用 gulp 縮小和重命名的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！