How to copy my build folder to remote server using gulp - node.js

Question

I am new to Gulp build tool.I recently starts using Gulp in my project.I done correctly the following things minified, compress,jsx file to js file.But My requirement is to send my build folder to my remote server which is located at another place.For this i am using gulp-scp npm to copy folder to remote server but it not copying folder to server.
The below code is gulpfile.js
var gulp = require('gulp'),
scp = require('gulp-scp2');
gulp.task('default', function() {
return gulp.src('src/*.js')
.pipe(scp({
host: '192.50.10.31',
username: 'applmgr',
password:'123456',
dest: '/Data/project'
}))
.on('error', function(err) {
console.log(err);
});
});
In this code i am copying all js files project in Data folder of remote server.
I run my gulpfile,js
c:\gulpproject>gulp
[10:19:36] Using gulpfile c:\gulpproject\gulpfile.js
[10:19:36] Starting 'default'...
coming like this and not showing any hing
can any help me why it is coming like this and in which place my code is wrong

Answer 1

Ok I solved this problem using gulp-rsync npm plugin.But In windows it you want copy file from your local to remote system for this you system support rsync and also contain ssh command.I tried this one in Linux and install rsync.
var gulp = require('gulp');
var uglify = require('gulp-uglify');
var react = require('gulp-react');
var rsync = require('gulp-rsync');
gulp.task('scripts', function() {
// Minify and copy all JavaScript (except vendor scripts)
// with sourcemaps all the way down
return gulp.src("src/*.jsx")
.pipe(react())
.pipe(uglify())
.pipe(gulp.dest('build/js'));
});
gulp.task('deploy',['scripts'], function() {
gulp.src('build/**')
.pipe(rsync({
root: 'build',
hostname: 'ramprasad#10.60.48.11',
destination: '/Data/rrv'
}));
gulp.src('build/**')
.pipe(rsync({
root: 'build',
hostname: 'appl#10.30.10.42',
destination: '/Data/rrv'
}));
});
// The default task (called when you run `gulp` from cli)
gulp.task('default', ['deploy']);

Answer 2

you can use powershell or cmd or batch:
var spawn = require("child_process").spawn,child;
child = spawn("powershell.exe",["c:\\temp\\helloworld.ps1"]);
child.stdout.on("data",function(data){
console.log("Powershell Data: " + data);
});
child.stderr.on("data",function(data){
console.log("Powershell Errors: " + data);
});
child.on("exit",function(){
console.log("Powershell Script finished");
});
child.stdin.end(); //end input