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Gulp process all files in directory

Tags:

gulp

I have css and js files in a directory (and subdirectories). I'm looking into different tools to compress the assets in all the directories. I'm trying to find a way to get gulp to compress all the files in those directories and save the compressed file in the same directory and name it with the following convention: [name].min.css or [name].min.js. So example.js would become example.min.js.

Is there a way to achieve this?

I've read the following on this:

http://gulpjs.com/plugins/

https://github.com/gulpjs/gulp/blob/master/docs/getting-started.md

https://github.com/gulpjs/gulp/blob/master/docs/API.md

like image 468
KVISH Avatar asked Feb 13 '16 05:02

KVISH


1 Answers

You usually don't want to generate the minified files in the same directory as the original files. You write all files that are generated by your build script to a single output directory. Some advantages of this approach are:

  • Makes it easier to clean the build and recreate everything from scratch: you just delete that one output folder.
  • You don't have to worry about generated files accidentally getting picked up by your build and getting processed again.

But since you asked, here's a solution that creates the minified files in the same directory as the original files. This creates a .min.css and .min.js file for every .css and .js file. All CSS files are assumed to be in a directory called css (or its subdirectories) and all JS files are assumed to be in a directory called js (or its subdirectories):

var gulp = require('gulp');
var rename = require('gulp-rename');
var cssnano = require('gulp-cssnano');
var uglify = require('gulp-uglify');

gulp.task('css', function () {
  return gulp.src([
      'css/**/*.css',
      '!css/**/*.min.css',
     ])
    .pipe(cssnano())
    .pipe(rename(function(path) {
      path.extname = ".min.css";
    }))
    .pipe(gulp.dest('css'));
});

gulp.task('js', function () {
  return gulp.src([
      'js/**/*.js',
      '!js/**/*.min.js',
     ])
    .pipe(uglify())
    .pipe(rename(function(path) {
      path.extname = ".min.js";
    }))
    .pipe(gulp.dest('js'));
});

gulp.task('default', ['css', 'js']);

Notice the negation pattern !css/**/*.min.css that is used to prevent the already minified CSS from getting minified again on the next build. Same for the JavaScript.

I used gulp-cssnano and gulp-uglify to minify the CSS and JS, but there's plenty of other options out there that can act as drop-in replacements.

like image 176
Sven Schoenung Avatar answered Nov 24 '22 13:11

Sven Schoenung