Send arguments to gulp task - terminal

Question

There are other questions on stackoverflow about this topic, but not the one I was hoping for.
Here is a simplified version of my gulpfile.js
var gulp = require('gulp');
var sass = require('gulp-sass');
gulp.task('css', function() {
gulp.src('site/scss/style.scss')
.pipe(sass())
.pipe(gulp.dest('assets/css'))
});
gulp.task('watch', function() {
gulp.watch('site/scss/**/*.scss', ['css']);
});
I can do this successfully:
gulp watch
Now I want to be able to do something like this in my terminal command line:
gulp watch my-domain-name
or
gulp watch my-other-domain
My gulpfile would in my mind look something like this:
var gulp = require('gulp');
var sass = require('gulp-sass');
gulp.task('css', function(name) {
gulp.src('site/scss/style.scss')
.pipe(sass())
.pipe(gulp.dest('assets/css'))
});
gulp.task('watch', function(name) {
gulp.watch('site/scss/**/*.scss', ['css', name]);
});
I try to send name around as a variable of my watch task. In my case it would be my-domain-name or my-other-domain depending on what I write in my terminal.
How can I send a parameter from the terminal to the watch task and then over to the css task?

Answer 1

There are a number of sources for passing parameters from the command line to gulp. See:
https://www.sitepoint.com/pass-parameters-gulp-tasks/
https://github.com/gulpjs/gulp/blob/master/docs/recipes/pass-arguments-from-cli.md
The official recipe uses
minimist an extremely popular package.
So with these approaches you would not be passing the command line args directly to the watch task, but rather to an object that can be accessed by any task.
And it look like gulp-param may do exactly what you want, directly injecting command line args into each task's callback function.