Combine gulp tasks to avoid creating the unnecessary files

Question

I'm using two separate gulp tasks to minify templates into js file (1) and concat all js file of the project to the only minified (2).

gulp.task('templates', function() { // RESULT of this task
    return gulp.src('templates/**/*.html')
        .pipe(dotify({
            root : 'templates'
        }))
        .pipe(concat('templates.js'))
        .pipe(gulp.dest('js'));
    });


gulp.task('minifyJs', ['templates'], function() {
    return gulp.src([
        'js/templates.js',
        'js/plugins/*.js',
        'js/*.js'
    ])
    .pipe(concat('scripts-all.js'))
});

The question is: am I able to avoid creating the templates.js file by processing the result from first task to the second one to concat it with the rest of js's?

score 0 · Accepted Answer · answered Oct 23 '15 at 12:31

Solution: addSrc should be used

return gulp.src('templates/**/*.html')
    .pipe(dotify({
        root : 'templates'
    }))
    .pipe(addSrc([
        'js/plugins/*.js',
        'js/common/*.js',
        'js/ui/*.js',
        'js/pages/*.js',
        'js/*.js'
    ]))
    .pipe(concat('scripts-all.js'))
    .pipe(gulp.dest('js/'));

Combine gulp tasks to avoid creating the unnecessary files

1 Answers1