CODEWITHSUNDEEP
javascriptgulp
Dhruv Balhara
Dhruv Balhara

Reputation: 347

Gulp.js - Not able to copy generated files from one folder to another

gulp.task('minify-css', function () {

   gulp.src(['src/test/test.css])
   .pipe(concat('test1.css'))
   .pipe(minifyCSS())
   .pipe(gulp.dest('src/source/'))
   .pipe(filesize())

});
gulp.task('copy-css',['minify-css'], function () {

   gulp.src('src/source/*.css')
   .pipe(gulp.dest('src/dest/'));

});

It seems that the first time I run 'gulp copy-css'

Starting 'minify-css'...
[18:54:39] Finished 'minify-css' after 61 ms
[18:54:39] Starting 'copy-css'...
[18:54:39] Finished 'copy-css' after 1.86 ms

but the copy operation doesn't execute probably because it executes even before the file to be copied is not generated

Even though I have mentioned minify-css as dependency for copy-css task, it seems it is not following that convention in this case.

When gulp copy-css is run another time, this time the file is copied because the file is already generated from previously executed command. But this would beat the purpose when the code is being used in production.

Upvotes: 4

Views: 1382

Answers (3)

harishr
harishr

Reputation: 18055

change both the tasks as below

gulp.task('minify-css', function () {
    return  gulp.src(['src/test/test.css])
   .pipe(concat('test1.css'))
   .pipe(minifyCSS())
   .pipe(gulp.dest('src/source/'))
   .pipe(filesize())    
});

gulp.task('copy-css',['minify-css'], function () {
   return gulp.src('src/source/*.css')
   .pipe(gulp.dest('src/dest/'));

});

add return so that next task runs after first one runs, else your copy-csss is running even before minify-css is finished.. hence the error...

Upvotes: 3

aludvigsen
aludvigsen

Reputation: 5981

Try with a callback in the on('finish') event:

gulp.task('minify-css', function (cb) {

   gulp.src(['src/test/test.css'])
    .pipe(concat('test1.css'))
    .pipe(minifyCSS())
    .pipe(gulp.dest('src/source/'))
    .pipe(filesize())
    .on('finish', function() {
        // All tasks should now be done
        return cb();
    })

});

gulp.task('copy-css',['minify-css'], function () {

   gulp.src('src/source/*.css')
     .pipe(gulp.dest('src/dest/'));

});

PS: You also go an syntax error in your gulp.src array at the very top, just a missing quote ' here gulp.src(['src/test/test.css])

Upvotes: 1

Antonio Pavicevac-Ortiz
Antonio Pavicevac-Ortiz

Reputation: 7739

I would check out this question, as it seems you might have to add return keyword before gulp.src(['src/test/test.css])...

Upvotes: 1

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