NPM clean modules

Is there a way to get npm to unbuild all the modules under node_modules? Something like npm rebuild that removes all build artifacts but doesn't rebuild them?

You can just delete the node_module directory

rm -rf node_modules/

You can take advantage of the 'npm cache' command which downloads the package tarball and unpacks it into the npm cache directory.

The source can then be copied in.

Using ideas gleaned from https://groups.google.com/forum/?fromgroups=#!topic/npm-/mwLuZZkHkfU I came up with the following node script. No warranties, YMMV, etcetera.

var fs = require('fs'),
path = require('path'),
exec = require('child_process').exec,
util = require('util');

var packageFileName = 'package.json';
var modulesDirName = 'node_modules';
var cacheDirectory = process.cwd();
var npmCacheAddMask = 'npm cache add %s@%s; echo %s';
var sourceDirMask = '%s/%s/%s/package';
var targetDirMask = '%s/node_modules/%s';

function deleteFolder(folder) {
    if (fs.existsSync(folder)) {
        var files = fs.readdirSync(folder);
        files.forEach(function(file) {
            file = folder + "/" + file;
            if (fs.lstatSync(file).isDirectory()) {
                deleteFolder(file);
            } else {
                fs.unlinkSync(file);
            }
        });
        fs.rmdirSync(folder);
    }
}

function downloadSource(folder) {
    var packageFile = path.join(folder, packageFileName);
    if (fs.existsSync(packageFile)) {
        var data = fs.readFileSync(packageFile);
        var package = JSON.parse(data);

        function getVersion(data) {
            var version = data.match(/-([^-]+)\.tgz/);
            return version[1];
        }

        var callback = function(error, stdout, stderr) {
            var dependency = stdout.trim();
            var version = getVersion(stderr);
            var sourceDir = util.format(sourceDirMask, cacheDirectory, dependency, version);
            var targetDir = util.format(targetDirMask, folder, dependency);
            var modulesDir = folder + '/' + modulesDirName;

            if (!fs.existsSync(modulesDir)) {
                fs.mkdirSync(modulesDir);
            }

            fs.renameSync(sourceDir, targetDir);
            deleteFolder(cacheDirectory + '/' + dependency);
            downloadSource(targetDir);
        };

        for (dependency in package.dependencies) {
            var version = package.dependencies[dependency];
            exec(util.format(npmCacheAddMask, dependency, version, dependency), callback);
        }
    }
}

if (!fs.existsSync(path.join(process.cwd(), packageFileName))) {
    console.log(util.format("Unable to find file '%s'.", packageFileName));
    process.exit();
}

deleteFolder(path.join(process.cwd(), modulesDirName));
process.env.npm_config_cache = cacheDirectory;
downloadSource(process.cwd());

In a word no.

In two, not yet.

There is, however, an open issue for a --no-build flag to npm install to perform an installation without building, which could be used to do what you're asking.

See this open issue.

I think you're looking for npm prune

npm prune [<name> [<name ...]]

This command removes "extraneous" packages. If a package name is provided, then only packages matching one of the supplied names are removed.

Extraneous packages are packages that are not listed on the parent package's dependencies list.

See the docs: https://docs.npmjs.com/cli/prune