Generate critical CSS for all .html pages with Gulp

Question

So, there is this npm package Google suggests to generate critical CSS:

var critical = require('critical');

gulp.task('critical', function (cb) {
    critical.generate({
        base: 'app/',
        src: 'index.html',
        dest: 'css/index.css'
    });
});

It is actually an awesome tool which works like a charm. The only disadvantage is that it only works for a single file in Gulp as opposed to other tasks that can be called with gulp.src('app/*.html') for all matching files.

So I decided to list all my pages in an array and iterate through them:

var pages = ['index', 'pricing'];

gulp.task('critical', function (cb) {
    for (var i = 0; i < pages.length; i++) {

        critical.generate({
            base: 'app/',
            src: pages[i] + '.html',
            dest: 'css/' + pages[i] + '.css'
        });
    }
});

which proved to be a working solution too (except some node.js memory issues).

The problem now is that I have to list the pages manually in the array. So I wonder if there is a way to generate such a list automatically with a function similar to gulp.src('app/*.html') eg to include all .html pages from a selected folder to generate critical CSS for?

Any help would be greatly appreciated!

Answer 1

You could do something like (pseudocode follows):

var foreach = require('gulp-foreach');

gulp.task('default', function () {

  return gulp.src('./yourPathTo/*.html')

          // treats each file in gulp.src as a separate stream
    .pipe(foreach(function (stream, file) {

         // see in next code for example of following
         var fileBase = file.path.stringOptoRemove Extension;

         critical.generate({
            base: 'app/',
            src: file.path,
            dest: 'css/' + fileBase + '.css'
        });
   }))
});

Using gulp-foreach .

Alternatively, glob-fs would be useful too and I include example code I wrote below that manipulates individual files from a stream and then writes them out with a new file name.

var gulp = require('gulp');
var glob = require('glob-fs')({ gitignore: true });
var html2text = require('html-to-text');
var fs = require('fs');
var path = require('path');  

gulp.task('default', function () {

  glob.readdirStream('build/*.html')
    .on('data', function (file) {
      console.log(file.path);

      html2text.fromFile(file.path, (err, text) => {
        if (err) return console.error(err);

        // strip off extension and parent directories
        var filename = path.basename(file.path, '.html');

        // I assume you want a.html text to be written to 'build/txt/a.txt'
        fs.writeFileSync(path.join('build/txt', filename  + '.txt'), text, 'utf8');
      });
    });
});

Answer 2

I have this working using gulp.src. only issue is that it takes a good 3 minutes to complete on sites with many html files and is susceptible to crashing if you are doing other things on your computer at the same time. I actually found this post by trying to find a more efficient way to do this.

gulp.task('criticalCSS', ['compile-html', 'compile-styles'], function() {
    return gulp.src(config.dest.root + '**/*.html')
    .pipe(critical({
        base: config.dest.root, // "app/" which is the root of my build folder
        inline: true, 
        css: config.dest.css + styleBundleFilename, // "app/css/mainCssFilename"
        dimensions: [{
            width: 320,
            height: 480
            },{
            width: 768,
            height: 1024
            },{
            width: 1280,
            height: 960
            }],
        ignore: ['font-face']
    }))
    .on('error', function(err) { log(err.message); })
    .pipe(gulp.dest(config.dest.root));
});