How do I use express.static() with a directory path based on the URL?

Question

I am making a service involving a system where the user gets a free subdomain, like archiebaer.blahblahblah.demo , and I have a function to get the site config file ( siteconf() ) that contains a key called theme. I want archiebaer.blahblahblah.demo/theme-static/style.css to use express.static() to serve a folder based on that theme key.

Eg. app.get('/theme-static', express.static("themes/ABC/theme-static")); where ABC is the theme name.

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Example Scenario: johnsmith and archiebaer are both users. John's config file looks something like this: {theme:'retro'} , and Archie's is {theme:'slate'} . You can get a user's config file from the express route's request parameter using siteconf(req) . When I go to /theme-static/style.css on Archie's website, I should get the file ~/projectfolder/themes/slate/style.css , and on John's: ~/projectfolder/themes/retro/style.css .

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I assume the code would look something like this:

app.get('/theme-static', function (req, res) {
   res.send(express.static('themes/' + siteconf(req).theme + '/theme-static/'));
});

Answer 1

Here was my solution:

app.get('/theme-static/*', function (req, res, next) {
  var relurl = req.url.replace("/theme-static/", "");
  if (relurl === '') {
    //This is so '/theme-static/' without any file after is treated as normal 404.
    next();
    return;
  }
  var filePath = "themes/" + siteconf(req).theme + "/theme-static/" + relurl;
  if (fs.existsSync(filePath)) {
    res.sendFile(path.join(__dirname, filePath));
  } else {
    res.status(404).send("Invalid File");
  }
});

Answer 2

You can simply use something like:

app.use("/theme-static", express.static(path.join(__dirname, "pathToFolder")));

If the file doesn't exist, express will handle the error for you.