gulp: passing dependent task return stream as parameter

Question

I'm trying to create two gulp tasks, and I'd like the second task to take the first one's output stream and keep applying plugins to it.

Can I pass the first task's return value to the second task?

The following doesn't work:

// first task to be run
gulp.task('concat', function() {
  // returning a value to signal this is sync
  return
    gulp.src(['./src/js/*.js'])
      .pipe(concat('app.js'))
      .pipe(gulp.dest('./src'));
};

// second task to be run
// adding dependency
gulp.task('minify', ['concat'], function(stream) {
  // trying to get first task's return stream
  // and continue applying more plugins on it
  stream
    .pipe(uglify())
    .pipe(rename({suffix: '.min'}))
    .pipe(gulp.dest('./dest'));
};

gulp.task('default', ['minify']);

Is there any way to do this?

Answer 1

you can't pass stream to other task. but you can use gulp-if module to skip some piped method depending on conditions.

var shouldMinify = (0 <= process.argv.indexOf('--uglify'));

gulp.task('script', function() {
  return gulp.src(['./src/js/*.js'])
      .pipe(concat('app.js'))
      .pipe(gulpif(shouldMinify, uglify())
      .pipe(gulpif(shouldMinify, rename({suffix: '.min'}))
      .pipe(gulp.dest('./dest'));
});

execute task like this to minify

gulp script --minify