gulp.task('scripts', function() {
return gulp.src(['!dev/js/vendor-list.js', 'dev/js/**/*.js'])
.pipe(concat('all.min.js'))
.pipe(gulp.dest('public'));
});
The above code is a sample of my gulp task using gulp-concat
. Somewhere in the pile is my app.js
file which I would like concatenated last. Is there a way to do it without having to list all the sub-folders in order? All other files are order agnostic. My app.js
is the only one order sensitive that I need concatenated last. I thought about doing two streams in a single task and then merging the output, but I was hoping for a simpler solution. Any thoughts?
Try this (from this question):
[
'!dev/js/vendor-list.js', // can probably be included in the statement below.
'/dev/js/!(app)*.js', // all files that end in .js EXCEPT app*.js
'/dev/js/app.js', // this should be concatenated last
]