Gulp function to print out less verbose console log message?

Question

I'm new to Gulp (or Grunt) tools. I created a function to print out the version value in my package.json file, like this

const {dest, parallel, series, src } = require('gulp');
const semver = require('semver');

const package_json = path.join(path.dirname(__filename),'package.json');
const pkg = JSON.parse(fs.readFileSync(package_json));
const version = semver.parse(pkg.version);

function get_version(cb) {
  console.log(version.version);
  cb();
}
exports.get_version = get_version

and I get

$ gulp get_version -f gulpfile-version.js 
[16:07:08] Using gulpfile ~/some_path/gulpfile-version.js
[16:07:08] Starting 'get_version'...
1.4.6
[16:07:08] Finished 'get_version' after 1.91 ms

How can I modify my function to be less verbose and only print out the 1.4.6 string, and not the other time-stamped lines?

Answer 1

You can use --silent flag in your command to suppress the gulp logs. Refer the doc https://github.com/gulpjs/gulp/blob/master/docs/CLI.md

Your example

$ gulp get_version -f gulpfile-version.js --silent