Execute gulp task only if flag is passed? -

current code:

var open = require('open'); var gulp = require("gulp"); var pkg = require("../../package.json");  //opens launchpage default browser gulp.task("open", function () {     open('http://localhost:' + pkg.webserverport + '/commonclient/launchpage.html'); }); 

right now, running gulp command starts server , bunch of other compilation tasks. how modify run "open" task if type in gulp -o or in command line?

you can inspecting process.argv, or looking env flag on process.env

gulp.task("open", function () {   if (process.argv.indexof('-o') > -1) {     open('http://localhost:' + pkg.webserverport + '/commonclient/launchpage.html');   } }); 

or env variable

gulp.task("open", function () {   if (processs.env.use_open) {     open('http://localhost:' + pkg.webserverport + '/commonclient/launchpage.html');   } });