How to omit devDependencies when copying node_modules? - gulp

Question

I'd like to copy only modules, which are important for application - those located inside dependencies in package.json. I'd like to omit those under devDependencies.
package.json
{
"dependencies": {
"express": "^4.13.4",
"log4js": "^0.6.33"
},
"devDependencies": {
"gulp": "^3.9.1",
"gulp-rename": "^1.2.2",
"gulp-typescript": "^2.12.1",
"typings": "^0.7.9",
"yargs": "^4.3.2"
}
}
gulpfile.js
gulp.task('copy_packages', function() {
gulp
.src('node_modules/**/*.*')
.pipe(gulp.dest('../release/node_modules'));
});
Is there any module or a smart way to distinguish which modules belongs to dependencies group and which to devDependencies?

Answer 1

Had the same problem...felt weird this was not considered when gulp was created.
My work around was using child_process to run npm install and specify a directory to put the node_modules directory with only the packages you need for your application.
e.g:
gulp.task('createDeployNodeModulesFolder', function(cb) {
spawn('"npm"', ['install', '--prefix', './dist/', 'package1', 'package2'], {stdio: 'inherit', shell: true}, function (err, stdout, stderr) {
console.log(stdout);
console.log(stderr);
})
.on('close', cb);
});
In your case you want only the production dependencies so you can likely use:
npm install --prefix /deployDir/ --only=prod
There will be some warnings complaining about no package.json..etc., but those are just warnings. If you really want to get rid of them, I guess you can just add a task to copy or create a package.json into the deploy directory before running npm install.

Answer 2

Node.js allows you to require() JSON files which will be returned as simple JavaScript objects. You can use that to only pass those modules to gulp.src() that appear under dependencies in your package.json file:
var gulp = require('gulp');
var packageJson = require('./package.json');
gulp.task('copy_packages', function() {
var modules = Object.keys(packageJson.dependencies);
var moduleFiles = modules.map(function(module) {
return 'node_modules/' + module + '/**/*.*';
});
return gulp.src(moduleFiles, { base: 'node_modules' })
.pipe(gulp.dest('../release/node_modules'));
});

Answer 3

use distize https://www.npmjs.com/distize .
npx distize --no-files -o {targetpath}