Gulp Tasks with Command Line Parameters

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Gulp Tasks with Command Line Parameters

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#1 by (5 votes)

5

Work with a simple task of Gulp , created to perform a deploy in the production environment. In a simplified way, the task is:

gulp.task('deploy-prod', function () {
    return gulp.src('dist/**')
        .pipe(sftp({
            host: '000.000.000.000',
            user: 'user',
            pass: 'pass',
            remotePath: '/var/www/html/meusite/',
            port: 22
        }))
        .pipe(gulp.dest('dist'));
});

I swipe with com

$ gulp deploy-prod

There is another environment (of homologation) that obviously has different addresses and credentials. For this, I have a task named deploy-homolog , whose logic is the same. This is just one case, among many others, where I end up having to repeat the code of my tasks because they present slightly different characteristics at the moment of development.

My question is: is there a way to parameterize the execution of these tasks via the command line? My intention was to roll

$ gulp deploy --prod

or

$ gulp deploy --homolog

and have only one task deploy , which would execute a routine based on this flag. It is possible? If so, how?

javascript gulpjs gulp

asked by anonymous 15.06.2016 / 15:38

1 answer

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score 5 · Accepted Answer

By default gulp does not accept parameters in this way, but you can use the yargs module to work with these arguments (there are several modules for this situation).

I will use a simple example but that can be used to solve your problem, follow the code:

gulpfile.js

var gulp = require('gulp')
    yargs = require('yargs'),
    args = yargs.argv;

gulp.task(
    'default',
    function () {
        if (args.prod) {
            console.log('Production tasks are running.');
        }
        if (args.dev) {
            console.log('Development tasks are running');
        }
    }
);

Now you can run the jobs in production with the command gulp --prod , or, the tasks in development with comando gulp --dev for example.

Editing

I'll add a second way to get the same result, but without using a new module / dependency. Here's how:

gulpfile.js

var gulp = require('gulp');

gulp.task(
  'deploy',
  function () {
    var production = process.argv.indexOf('--production') !== -1;

    if (production) {
      console.log('Running production tasks.');
    }
    if (! production) {
      console.log('Running development tasks.');
    }
  }
);

Now you can run gulp deploy --production to run production tasks, or run gulp deploy , to run development tasks.