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nmp install --save-dev gulp-environments
Should be:
npm install --save-dev gulp-environments

I wanted to create 2 environments, based on my process.env.HOST variable, which should be either "live" or "dev".

var environments = require('gulp-environments');
var liveOnly = environments.make('live');
var devOnly = environments.make('dev');
environments.current(process.env.HOST);
console.log(liveOnly?true:false,devOnly?true:false)

This is on my dev environment, and I know process.env.HOST resolves to "dev". But in the above example, the console logs true true, which according to your docs indicates that both environments are active? Maybe I'm misunderstanding what "active" means, but I took it to mean it's set as the current task.

Also if I use console.log(liveOnly(),devOnly()) i instead get false false, not sure what that's about.

Given the line current(make(process.env.NODE_ENV || argv.env || "development")); if NODE_ENV is set to anything truthy it'll use that regardless of any other method (as the or is short-circuited) of setting the environment, and otherwise it'll fall to arg.env in the same way, and so forth. Make this clear in the documentation. Also, if NODE_ENV is set to "development" then not even gulp.task('set-prod', production.task); or environments.current(production); will override it

Plus, I think it should be reversed, command line should override NODE_ENV or code, and code should override NODE_ENV.

Thanks for the module!

I have files that are supposed to be compiled to temp folder if the NODE_ENV value is development; otherwise, to app folder.

I've tried these 4 ways but all failed:

// the $ sign is gulpLoadPlugins()
const isDev = $.environments.development;
const isProd = $.environments.production;

1. As what the docs instructed
   .pipe(isDev() ? gulp.dest('temp') : gulp.dest('app'))

2. With gulp-if
.pipe($.if(isDev(), gulp.dest('temp'), gulp.dest('app')))

3. Processing the condition inside the gulp.dest() method
   .pipe(gulp.dest(isDev() ? 'temp' : 'app'))

4. Both condition piped
   .pipe(isDev(gulp.dest('temp')))
.pipe(isProd(gulp.dest('app')))

THE RESULT:
Both NODE_ENV values compile files to the same folder.

Removing the parentheses after isDev doesn't solve the problem either.

This would save a lot of typing in the CLI when in development. I would argue it's also the expected behavior.

If I have two tasks written in the default Gulp 4 export mode (task() is deprecated), in example:

exports.development = parallel(css, js, img, fonts, icons);
exports.production = parallel(css, js, img, fonts, icons);

how can I set the environment under which each of these exported task should run form inside the gulpfile (I don't want to use CLI flags)?

The readme says to create a task that sets the environment and run it before the others, what is the correct way to do it by using the default Gulp 4 function definition of tasks, like in the example below?

function setDev(cb) {
  // body omitted
  cb();
}

function setProd(cb) {
  // body omitted
  cb();
}

I would love to have the ability to do an inline check for production or staging in my gulp tasks, maybe something like the following:

...
    .pipe(environments.in(['production', 'staging'], uglify()))
...or...
    .pipe(environments.not(['production', 'staging'], uglify()))
...

Is there currently a supported way of doing this?

I want to be able to output the currently defined environment as a string; environments.current() should return "development"