How to copy files based on environment on gulp - angularjs

Question

I have folder with files for different build environment like production,stage, and test.
example:
src/config.prod.js
src/config.stage.js
src/config.test.js
what I want is to copy config file based on environment I got, for getting environment name from command i using following code:
var nopt = require('nopt')
, knownOpts = {
"env" : [String, null]
}
, shortHands = {
"test" : ["--env", "test"]
, "dev" : ["--dev", "dev"]
, "stage" : ["--env", "stage"]
, "prod" : ["--env", "prod"]
};
var flags = nopt(knownOpts, shortHands, process.argv, 2);
and when I hit command
gulp build --env dev
I am getting the environment name, now what I want is to copy config file to dist folder(build) based on environment. I have this task for copy file but it copy all files as I don't know how to filter it out.
gulp.task('copyConfig', function(){
gulp.src(['src/*.js'])
.pipe(gulp.dest('dist/'))
})
I am new to gulp, if some one has any suggestion. please help.

Answer 1

new configs directory where you mv your 3, existing configs (prod, stage, test) ... mkdir configs
Then , 2 related changes to gulp to incorporate build.ENV and a copy.config task that knows about your 3 diff config files...
var settings = {
/*
* Environment development | production
* match './configs' files of same name
*/
environment : process.env.NODE_ENV || 'development',
/*
* Where is our config folder?
*/
configFolder : 'configs',
/*
* Where is our code?
*/
srcFolder : 'app/scripts',
/*
* Where are we building to?
*/
buildFolder : 'dist',
};
/**
* Config Task
*
* Get the configuration file (dev or prod), rename it
* and move it to be built.
*/
gulp.task('config', function() {
return gulp.src(settings.configFolder + '/' + settings.environment + '.js')
.pipe(rename('config.js'))
.pipe(gulp.dest(settings.srcFolder));
});