CODEWITHSUNDEEP
gulp
HXH
HXH

Reputation: 1663

About gulp task dependences

I write gulpfile.js like:

...

gulp.task('styles',function(){
  gulp.src('./src/sass/*.scss')
    .pipe(sass())
    .pipe(autoprefixer('last 2 version'))
    .pipe(concat('tmp.css'))
    .pipe(gulp.dest('./build/css'))
    .pipe(notify('Styles task complete'))
})

gulp.task('css',['styles'],function(cb){
  del(['./build/css/tmp.css'],cb)
})
.....

When I run gulp css,the styles task will first be execute, so ./build/css/tmp.css will be removed.

But the ./build/css/tmp.css is still exist.That is mean,when the css task be executed,the styles task has not yet be execute.

Why the dependences not work? and how can I make sure the './build/css/tmp.css' has exist before the css task be executed?

Upvotes: 1

Views: 47

Answers (1)

Pogrindis
Pogrindis

Reputation: 8091

I have seen this issue intermittently working without the returns in dependencies.

You need to return the actual task so gulp knows when it's complete.

So this should become : 

gulp.task('styles',function(){
  return gulp.src('./src/sass/*.scss')
    .pipe(sass())
    .pipe(autoprefixer('last 2 version'))
    .pipe(concat('tmp.css'))
    .pipe(gulp.dest('./build/css'))
    .pipe(notify('Styles task complete'))
});

gulp.task('css',['styles'],function(cb){
  del(['./build/css/tmp.css'],cb)
})

This will then wait for styles to complete (as a dependency returned) and then continue.

Upvotes: 1

Related Questions