How do I set up a nodejs server that gives the client the correct file by checking the url?

Question

I am trying to write a node js server that gives back whatever the user requested, if the resource is in the file system.

For example, if the request URL were /index.html , it would try to find a file called 'index.html' in the root directory and respond with a stream to that file. If the request were /myscript.js , it would do the same, finding a file called myscript.js and piping that to the response.

Here is what I have so far:

var http = require("http");
var fs = require("fs");
var port = process.env.PORT || 3000;
http.createServer(function (request, response) {
    if (request.method == "GET") {
        console.log(request.url);
        if (request.url == "/") { // if the url is "/", respond with the home page
            response.writeHead(200, {"Content-Type": "text/html"});
            fs.createReadStream("index.html").pipe(response);
        } else if (fs.existsSync("." + request.url)) {
            response.writeHead(200/*, {"Content-Type": request.headers['content-type']}*/);
            fs.createReadStream("." + request.url).pipe(response);
        } else {
            response.writeHead(404, {"Content-Type": "text/plain"});
            response.end("404 Not Found");
        }
    } else {
        response.writeHead(404, {"Content-Type": "text/plain"});
        response.end("404 Not Found");
    }
}).listen(port);

// Console will print the message
console.log('Server running at http://127.0.0.1:' + port +'/');

There are a few things that I don't like about this code:

• Apparently fs thinks that whenever the path starts with / , the file does not exist, so I must add a . before the file path. See line 10 and 12. Will this always work? I feel like this is a really bad trick to work around the problem.
• I don't know the content type of the requested file (line 11). I searched online and found a lot of ways to do this with other modules that I have to install using npm. Is there something in node that will figure out the content type?

Answer 1

Addressing your two bullet points:

• It makes sense that you would need a . when using fs because fs will look at the root directory of your system otherwise. Remember, the system root is different than your server root.
• One idea is to use path.extname and then a switch statement to set the content type like so.

The code below is taken from an blog post I wrote on how to set up a simple static server using Node without Express:

let filePath = `.${request.url}`; // ./path/to/file.ext
let ext = path.extname(filePath); // .ext
let contentType = 'application/octet-stream'; // default content type

if(ext === '') {

    filePath = './index.html'; // serve index.html
    ext = '.html';
}

switch (ext) {

    case '.html':
        contentType = 'text/html';
        break;

    case '.css':
        contentType = 'text/css';
        break;

    case 'js':
        contentType = 'text/javascript';
}