uglifyjs

I have a directory like below: /folder/b.js /folder/jQuery.js /folder/a.js /folder/sub/c.js I want to minify all these js files in one js file in order : jQuery.js -> a.js -> b.js -> c.js Q: 1.How can I do it via grunt-contrib-uglify?(In fact, there are lots of files, it is impractical to specify all source filepaths individually) 2.btw, How can I get unminified files when debug and get minified single file when release and no need to change script tag in html(and how to write the script tag)? Good questions! 1) Uglify will reorder the functions in the destination file so that function

What's the difference between concat, uglify, and minify tasks in grunt? I set up an uglify task for all of my site's javascript tasks, and it seemed to both minify and concatenate them. Grunt's site has a great description for how to configure each task, but it doesn't seem to explain what each task actually does. diclophis Concatenation is just appending all of the static files into one large file . Minification is just removing unnecesary whitespace and redundant / optional tokens like curlys and semicolons, and can be reversed by using a linter. Uglification is the act of transforming the

Is it possible to compress multiple files with UglifyJS? Something like... uglifyjs -o app.build.js appfile1.js appfile2.js ... Also, I am running Uglify via NodeJS in Windows Command Prompt Actually what you want to do (trick it into thinking its just 1 file) is just cat it Linux cat file1.js file2.js file3.js file4.js | uglifyjs -o files.min.js Windows (untested) type file1.js file2.js > uglifyjs -o files.min.js OR type file1.js file2.js > merged.files.js uglifyjs -o merged.files.js For future readers of this question, w/ UglifyJS2, this is trivial now... uglifyjs file1.js file2.js -o foo

I tried to uglify output of Browserify in Gulp, but it doesn't work. gulpfile.js var browserify = require('browserify'); var gulp = require('gulp'); var uglify = require('gulp-uglify'); var source = require('vinyl-source-stream'); gulp.task('browserify', function() { return browserify('./source/scripts/app.js') .bundle() .pipe(source('bundle.js')) .pipe(uglify()) // ??? .pipe(gulp.dest('./build/scripts')); }); As I understand I cannot make it in steps as below. Do I need to make in one pipe to preserve the sequence? gulp.task('browserify', function() { return browserify('./source/scripts/app

I have this node.js code that tries to minify and combine multiple js files to a single js file. var concat = require('gulp-concat'); var gulp = require('gulp'); gulp.task('scripts', function() { //gulp.src(['./lib/file3.js', './lib/file1.js', './lib/file2.js']) gulp.src(['./js/*.js']) .pipe(concat('all.js')) .pipe(uglify()) .pipe(gulp.dest('./dist/')) }); All my js files are located in js folder. My node.js file is above the js folder. I am expecting the single minified file to appear in dist folder. I see nothing and get no error message when I run the code. What could have gone wrong?

I'd like to know if there is a way to uglify only one file in command line using r.js (RequireJS Optimizer) which is already installed in my computer. Like we can minify a css file using node r.js cssIn="" out="" ... I'm actually working from a computer without internet connection, I'm not able to install something else on it (especially using npm) Maybe is there a possibility to download an uglifyjs package from another computer including all dependencies ready to be installed on mine? I didn't find something like that though... Thanks Copy uglify from another computer Since you mentioned