Proxy with express.js

To avoid same-domain AJAX issues, I want my node.js web server to forward all requests from URL /api/BLABLA to another server, for example other_domain.com:3000/BLABLA, and return to user the same thing that this remote server returned, transparently.

All other URLs (beside /api/*) are to be served directly, no proxying.

How do I achieve this with node.js + express.js? Can you give a simple code example?

(both the web server and the remote 3000 server are under my control, both running node.js with express.js)

So far I found this https://github.com/http-party/node-http-proxy , but reading the documentation there didn't make me any wiser. I ended up with

var proxy = new httpProxy.RoutingProxy();
app.all("/api/*", function(req, res) {
    console.log("old request url " + req.url)
    req.url = '/' + req.url.split('/').slice(2).join('/'); // remove the '/api' part
    console.log("new request url " + req.url)
    proxy.proxyRequest(req, res, {
        host: "other_domain.com",
        port: 3000
    });
});

but nothing is returned to the original web server (or to the end user), so no luck.

request has been deprecated as of February 2020, I'll leave the answer below for historical reasons, but please consider moving to an alternative listed in this issue.

Archive

I did something similar but I used request instead:

var request = require('request');
app.get('/', function(req,res) {
  //modify the url in any way you want
  var newurl = 'http://google.com/';
  request(newurl).pipe(res);
});

I hope this helps, took me a while to realize that I could do this :)

I found a shorter and very straightforward solution which works seamlessly, and with authentication as well, using express-http-proxy:

const url = require('url');
const proxy = require('express-http-proxy');

// New hostname+path as specified by question:
const apiProxy = proxy('other_domain.com:3000/BLABLA', {
    proxyReqPathResolver: req => url.parse(req.baseUrl).path
});

And then simply:

app.use('/api/*', apiProxy);

Note: as mentioned by @MaxPRafferty, use req.originalUrl in place of baseUrl to preserve the querystring:

    forwardPath: req => url.parse(req.baseUrl).path

Update: As mentioned by Andrew (thank you!), there's a ready-made solution using the same principle:

npm i --save http-proxy-middleware

And then:

const proxy = require('http-proxy-middleware')
var apiProxy = proxy('/api', {target: 'http://www.example.org/api'});
app.use(apiProxy)

Documentation: http-proxy-middleware on Github

I know I'm late to join this party, but I hope this helps someone.

You want to use http.request to create a similar request to the remote API and return its response.

Something like this:

const http = require('http');
// or use import http from 'http';


/* your app config here */

app.post('/api/BLABLA', (oreq, ores) => {
  const options = {
    // host to forward to
    host: 'www.google.com',
    // port to forward to
    port: 80,
    // path to forward to
    path: '/api/BLABLA',
    // request method
    method: 'POST',
    // headers to send
    headers: oreq.headers,
  };

  const creq = http
    .request(options, pres => {
      // set encoding
      pres.setEncoding('utf8');

      // set http status code based on proxied response
      ores.writeHead(pres.statusCode);

      // wait for data
      pres.on('data', chunk => {
        ores.write(chunk);
      });

      pres.on('close', () => {
        // closed, let's end client request as well
        ores.end();
      });

      pres.on('end', () => {
        // finished, let's finish client request as well
        ores.end();
      });
    })
    .on('error', e => {
      // we got an error
      console.log(e.message);
      try {
        // attempt to set error message and http status
        ores.writeHead(500);
        ores.write(e.message);
      } catch (e) {
        // ignore
      }
      ores.end();
    });

  creq.end();
});

Notice: I haven't really tried the above, so it might contain parse errors hopefully this will give you a hint as to how to get it to work.