Gulp uglify output min.js

Question

The js file being compressed is output with the same filename.

gulp.task('compressjs', function () {
  gulp.src('app/**/*.js')
    .pipe(uglify())
    .pipe(gulp.dest('dist'));
});

How can I have it output the file with min.js at the back?

Preview · Accepted Answer · 2016-08-19T08:42:44.597

129

You can use the suffix or extname parameters of gulp-rename like:

gulp.task('compressjs', function () {
  gulp.src('app/**/*.js')
    .pipe(uglify())
    .pipe(rename({ suffix: '.min' }))
    .pipe(gulp.dest('dist'))
})

edited Aug 19 '16 at 08:42

answered Jan 13 '15 at 08:40

Preview

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How would you update the html script tags to reflect the name change? I would like a gulp build that minified javascript files for production but not for development builds. – Nathan Reese Jun 30 '15 at 22:44
@NathanReese You could use a plugin like *gulp-rev-all* to do that – Preview Jun 30 '15 at 22:46
7

another (cleaner) way to do it is `.pipe(rename({ suffix: '.min' }))` – intellion Jan 28 '16 at 13:52
1

@intellion You're right, mind if I change it in the answer? – Preview Jan 28 '16 at 14:07
7

@Aperçu absolutely not :) – intellion Jan 28 '16 at 14:09
1

It's terrible that `gulp-clean-css` and `gulp-uglify` don't do this automagically. But this solution worked like charm. Thanks. – Sampgun Mar 19 '19 at 14:11
Don't forget to npm install gulp-rename and require it. – Jony Mar 21 '19 at 04:44

Gulp uglify output min.js

1 Answers1