gulp 将所有 css 文件缩小为单个文件 [英] gulp minify all css files to a single file

问题描述

我有如下 gulp 任务脚本，

I have gulp task script as following,

// loads various gulp modules
var gulp = require('gulp');
var concat = require('gulp-concat');
var minifyCSS = require('gulp-minify-css');
var autoprefixer = require('gulp-autoprefixer');
var rename = require('gulp-rename');

// create task
gulp.task('css', function(){
    gulp.src('src/css/**/*.css')
        .pipe(minifyCSS())
        .pipe(rename('style.min.css'))
        .pipe(autoprefixer('last 2 version', 'safari 5', 'ie 8', 'ie 9'))
        .pipe(gulp.dest('dist/css'))
});

如何将 src/css 中的所有 css 文件压缩为 dist/css/style.min.css 的单个文件?

How to minify all the css files from src/css to a single file as dist/css/style.min.css?

推荐答案

你的 gulp 任务缺少 concat 管道.

Your gulp task is missing the concat pipe.

gulp.src('src/css/**/*.css')
    .pipe(minifyCSS())
    .pipe(autoprefixer('last 2 version', 'safari 5', 'ie 8', 'ie 9'))
    .pipe(concat('style.min.css'))
    .pipe(gulp.dest('dist/css'))

这是一个关于使用 gulp 构建的非常好的教程:

Here is a very good tutorial about building with gulp:

http://www.smashingmagazine.com/2014/06/11/building-with-gulp/

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