javascript - Call task by default in gulp with flag -


someone can me? have next gulpfile...

    var          gulp                = require('gulp')         sass                = require('gulp-sass'),         cleancss            = require('gulp-clean-css'),         sourcemaps          = require('gulp-sourcemaps'),         rename              = require('gulp-rename'),         gulpngconfig        = require('gulp-ng-config'),         browsersync         = require('browser-sync').create(),         gutil               = require('gulp-util'),         environmentconfig   = require('./enviroment.json');     //enviroment configuration  gulp.task('setenviroment', function () {         gulp.src('enviroment.json')     .pipe(gulpngconfig('originacion.constants', {         environment: 'env.development'     }))     .pipe(rename('constants.js'))     .pipe(gulp.dest('js',{         overwrite:true     })) }); gulp.task('browser-sync', function() {     browsersync.init({         server: environmentconfig.path_app,         port: gutil.env.mod == 'development' ? environmentconfig.env.development.port : environmentconfig.env.production.port     }); });

if need call browser-sync task, use:

gulp browser-sync --mod production

i want call 2 tasks in gulp task, this:

gulp.task('init', ['setenviroment', 'browser-sync']);

my ask is: can call task default flag?... example

gulp.task('init', ['setenviroment', 'browser-sync:development']);

for call browser-sync flag --mod production default


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