how do I run a gulp task from two or more other tasks and pass the pipe through

Question

This must be obvious but I can't find it. I want to preprocess my stylus/coffee files with a watcher in the dev environment and in production with a build task (isn't that common to all of us?) and also run a few more minification and uglification steps in production but I want to share the pipe steps common to both dev and production for DRY

The problem is that when I run the task which watches the files, the task which preprocesses does that to all the files since it has its own gulp.src statement which includes all stylus files.

How do I avoid compiling all files on watching while still keeping the compile task separate. Thanks

paths = {
    jade: ['www/**/*.jade']
  };

  gulp.task('jade', function() {
    return gulp.src(paths.jade).pipe(jade({
      pretty: true
    })).pipe(gulp.dest('www/')).pipe(browserSync.stream());
  });



 gulp.task('serve', ['jade', 'coffee'], function() {
    browserSync.init({
      server: './www'
    });
    watch(paths.jade, function() {
      return gulp.start(['jade']);
    });
    return gulp.watch('www/**/*.coffee', ['coffee']);
  });

Answer 1

One important thing in Gulp is not to duplicate pipelines. If you want to process your stylus files, it has to be the one and only stylus pipe. If you want to execute different steps in your pipe however, you have multiple choices. One that I would suggest would be a noop() function in conjunction with a selection function:

var through = require('through2'); // Gulp's stream engine

/** creates an empty pipeline step **/
function noop() {
  return through.obj();
}

/** the isProd variable denotes if we are in
  production mode. If so, we execute the task.
  If not, we pass it through an empty step
  **/
function prod(task) {
  if(isProd) {
    return task;
  } else {
    return noop();
  }
}

gulp.task('stylus', function() {
  return gulp.src(path.styles)
    .pipe(stylus())
    .pipe(prod(minifyCss())) // We just minify in production mode
    .pipe(gulp.dest(path.whatever))
})

As for the incremental builds (building just the changed files with every iteration), the best way would be to get on the gulp-cached plugin:

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

gulp.task('stylus', function() {
  return gulp.src(path.styles)
    .pipe(cached('styles')) // we just pass through the files that have changed
    .pipe(stylus())
    .pipe(prod(minifyCss()))
    .pipe(gulp.dest(path.whatever))
})

This plugin will check if the contents have changed with each iteration you have done.

I spend a whole chapter on Gulp for different environments in my book, and I found those to be the most suitable ones. For more information on incremental builds, you can also check on my article on that (includes Gulp4): http://fettblog.eu/gulp-4-incremental-builds/