How to include scripts located inside the node_modules folder?

Solution 1:

Usually, you don't want to expose any of your internal paths for how your server is structured to the outside world. What you can is make a /scripts static route in your server that fetches its files from whatever directory they happen to reside in. So, if your files are in "./node_modules/bootstrap/dist/". Then, the script tag in your pages just looks like this:

<script src="/scripts/bootstrap.min.js"></script>

If you were using express with nodejs, a static route is as simple as this:

app.use('/scripts', express.static(__dirname + '/node_modules/bootstrap/dist/'));

Then, any browser requests from /scripts/xxx.js will automatically be fetched from your dist directory at __dirname + /node_modules/bootstrap/dist/xxx.js.

Note: Newer versions of NPM put more things at the top level, not nested so deep so if you are using a newer version of NPM, then the path names will be different than indicated in the OP's question and in the current answer. But, the concept is still the same. You find out where the files are physically located on your server drive and you make an app.use() with express.static() to make a pseudo-path to those files so you aren't exposing the actual server file system organization to the client.

If you don't want to make a static route like this, then you're probably better off just copying the public scripts to a path that your web server does treat as /scripts or whatever top level designation you want to use. Usually, you can make this copying part of your build/deployment process.

If you want to make just one particular file public in a directory and not everything found in that directory with it, then you can manually create individual routes for each file rather than use express.static() such as:

<script src="/bootstrap.min.js"></script>

And the code to create a route for that

app.get('/bootstrap.min.js', function(req, res) {
    res.sendFile(__dirname + '/node_modules/bootstrap/dist/bootstrap.min.js');
});

Or, if you want to still delineate routes for scripts with /scripts, you could do this:

<script src="/scripts/bootstrap.min.js"></script>

And the code to create a route for that

app.get('/scripts/bootstrap.min.js', function(req, res) {
    res.sendFile(__dirname + '/node_modules/bootstrap/dist/bootstrap.min.js');
});

Solution 2:

I would use the path npm module and then do something like this:

var path = require('path');
app.use('/scripts', express.static(path.join(__dirname, 'node_modules/bootstrap/dist')));

IMPORTANT: we use path.join to make paths joining using system agnostic way, i.e. on windows and unix we have different path separators (/ and )

Solution 3:

As mentioned by jfriend00 you should not expose your server structure. You could copy your project dependency files to something like public/scripts. You can do this very easily with dep-linker like this:

var DepLinker = require('dep-linker');
DepLinker.copyDependenciesTo('./public/scripts')
// Done

Solution 4:

If you want a quick and easy solution (and you have gulp installed).

In my gulpfile.js I run a simple copy paste task that puts any files I might need into ./public/modules/ directory.

gulp.task('modules', function() {
    sources = [
      './node_modules/prismjs/prism.js',
      './node_modules/prismjs/themes/prism-dark.css',
    ]
    gulp.src( sources ).pipe(gulp.dest('./public/modules/'));
});

gulp.task('copy-modules', ['modules']);

The downside to this is that it isn't automated. However, if all you need is a few scripts and styles copied over (and kept in a list), this should do the job.