Can gulp-uglify strip out console.log statements? - gulp

Question

As the question states. I know gulp-uglify can uglify with a simple:
gulp.src('filename')
.pipe(uglify())
Is there a way to tell it to strip out console.log statements too?

Answer 1

Yes there is!
As gulp-uglifyjs documentation mentions you can pass extra options ( gulp-uglifyjs documentation):
uglify([filename], [options])
All available options can be found on the compressor UglifyJS documentation page. From that 'drop_console: true' should help:
uglify([filename], {
compress: {
drop_console: true
}
})

Answer 2

Much better still: you can use the specialized gulp plugin: gulp-strip-debug. It doesn't only strip out console statements, but also alerts and debugger statements.
Strip console, alert, and debugger statements from JavaScript code with strip-debug
install it using:
npm install --save-dev gulp-strip-debug
and use it like:
var gulp = require('gulp');
var stripDebug = require('gulp-strip-debug');
gulp.task('default', function () {
return gulp.src('src/app.js')
.pipe(stripDebug())
.pipe(gulp.dest('dist'));
});

Answer 3

With gulp-uglify (not gulp-uglifyjs) you can remove arbitrary code by using a strategy similar to compilations constants with the compression configuration (Link goes to gulp-uglifyjs, but should be the same for gulp-uglify):
.pipe(uglify({
compress: {
global_defs: {
"DEBUG": false
}
}
}))
And then in your code write something like
if (DEBUG)
{
console.log("Application started...");
//...
}
if you set DEBUG in global_defs to true, gulp-uglify will remove the conditional but leave the inner branch. In case DEBUG is set to false, it will remove both.
drop_console as in Fill's answer does not seem to be supported in gulp-uglify.