Is it possible to pass a flag to Gulp to have it run tasks in different ways?

Question

Normally in Gulp tasks look like this:

gulp.task('my-task', function() {     return gulp.src(options.SCSS_SOURCE)         .pipe(sass({style:'nested'}))         .pipe(autoprefixer('last 10 version'))         .pipe(concat('style.css'))         .pipe(gulp.dest(options.SCSS_DEST)); }); 

Is it possible to pass a command line flag to gulp (that's not a task) and have it run tasks conditionally based on that? For instance

$ gulp my-task -a 1 

And then in my gulpfile.js:

gulp.task('my-task', function() {         if (a == 1) {             var source = options.SCSS_SOURCE;         } else {             var source = options.OTHER_SOURCE;         }         return gulp.src(source)             .pipe(sass({style:'nested'}))             .pipe(autoprefixer('last 10 version'))             .pipe(concat('style.css'))             .pipe(gulp.dest(options.SCSS_DEST)); }); 

Answer 1

Gulp doesn't offer any kind of util for that, but you can use one of the many command args parsers. I like yargs. Should be:

var argv = require('yargs').argv;  gulp.task('my-task', function() {     return gulp.src(argv.a == 1 ? options.SCSS_SOURCE : options.OTHER_SOURCE)         .pipe(sass({style:'nested'}))         .pipe(autoprefixer('last 10 version'))         .pipe(concat('style.css'))         .pipe(gulp.dest(options.SCSS_DEST)); }); 

You can also combine it with gulp-if to conditionally pipe the stream, very useful for dev vs. prod building:

var argv = require('yargs').argv,     gulpif = require('gulp-if'),     rename = require('gulp-rename'),     uglify = require('gulp-uglify');  gulp.task('my-js-task', function() {   gulp.src('src/**/*.js')     .pipe(concat('out.js'))     .pipe(gulpif(argv.production, uglify()))     .pipe(gulpif(argv.production, rename({suffix: '.min'})))     .pipe(gulp.dest('dist/')); }); 

And call with gulp my-js-task or gulp my-js-task --production.