I want to minify some JS files with Gulp, but can't seem to get control over the process. I want only the minified version in the destination, and am currently getting copies of the originals as well.

I'm thinking I may need the rename package, but am not sure how to use it for this task - I would presumably need some kind of variable to hold the current file name for each script.

Any help much appreciated. The code is below:

var gulp = require( 'gulp' );
var minify = require( 'gulp-minify' );
var rename = require( 'gulp-rename' );

//script paths
var jsFiles = 'js/**/*.js',
    jsDest = 'js/dist/';

gulp.task('scripts', function() {
    return gulp.src(jsFiles)
        .pipe(minify())
        .pipe(gulp.dest(jsDest));
});

I want to minify some JS files with Gulp, but can't seem to get control over the process. I want only the minified version in the destination, and am currently getting copies of the originals as well.

I'm thinking I may need the rename package, but am not sure how to use it for this task - I would presumably need some kind of variable to hold the current file name for each script.

Any help much appreciated. The code is below:

var gulp = require( 'gulp' );
var minify = require( 'gulp-minify' );
var rename = require( 'gulp-rename' );

//script paths
var jsFiles = 'js/**/*.js',
    jsDest = 'js/dist/';

gulp.task('scripts', function() {
    return gulp.src(jsFiles)
        .pipe(minify())
        .pipe(gulp.dest(jsDest));
});

• javascript
• gulp

• The minified files are written under js/**/*.js too ???? – Alejandro Teixeira Muñoz Commented Apr 18, 2018 at 16:12

1 Answer 1

Set the noSource config like the following

  .pipe(minify({noSource: true})