CSS缩小和重命名与gulp [英] CSS minify and rename with gulp

问题描述

我有一个变量，例如

var files = {
    'foo.css': 'foo.min.css',
    'bar.css': 'bar.min.css',
};

我想要gulp为我做的是 minify 文件，然后重命名。

What I want the gulp to do for me is to minify the files and then rename for me.

但是任务目前写为档案）

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('./'))
});

如何重写，使其适用于我的变量 files

How to rewrite so it work with my variable files defined above?

推荐答案

您应该能够使用Glob选择src所需的任何文件，在一个对象，这应该简化你的任务。此外，如果你想将css文件缩小到单独的文件，你不需要将它们连接起来。

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() {

});

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