How to concat/uglify files from bower using gulp while ignoring .min files?

Question

I am using bower and wiredep. I want to have a gulp task to combine and uglify all files into a single file... "allmyjs.min.js"

My problem is with libraries like "angular" where within the bower_components/angular directory, there are multiple js files:

angular.js
angular.min.js
index.js

All I want to pick up is the non-minified file so I can concat all my js and minify them, but how do I specify a rule in my list of files that can ignore the .min files?

Right now I have rules like:

"myjs/js/*.js"
"bower_components/angular/angular.js" <- this works, but I have to list out every single dependency which is a nightmare
//"bower_components/angular/*.js" <- this does not work because it picks up everything

I know when I do a wiredep, it looks like bower is smart enough to pick out the specific "JS" files that it injects into my HTML. Is there a way that I can do something in my gulpfile that is smart enough to read the dependencies in bower and include the correct/associated js files without me having to list out every single dependency: (e.g. "bower_comonents/angular/angular.js") for every single bower dependency?

Possible duplicate of [Excluding files/directories from Gulp task](http://stackoverflow.com/questions/23384239/excluding-files-directories-from-gulp-task) — Sergio Ivanuzzo, Nov 18 '15 at 21:23

score 1 · Accepted Answer · answered Nov 20 '15 at 23:51

In gulp.src a glob that begins with '!' excludes matching files from the glob results up to that point

var jsFiles = './js/**/*.js';
var minJsFiles = './js/**/*.min.js';

gulp.task('minify', function(){
    return gulp.src([jsFiles, '!' + minJsFiles])
    .pipe(doSomething()); // min files or some other stuff here
});

How to concat/uglify files from bower using gulp while ignoring .min files?

1 Answers1