How to programmatically send a 404 response with Express/Node?

Javascriptnode.jsExpressHttp Status-Codes

Javascript Problem Overview


I want to simulate a 404 error on my Express/Node server. How can I do that?

Javascript Solutions


Solution 1 - Javascript

Since Express 4.0, there's a dedicated sendStatus function:

res.sendStatus(404);

If you're using an earlier version of Express, use the status function instead.

res.status(404).send('Not found');

Solution 2 - Javascript

Updated Answer for Express 4.x

Rather than using res.send(404) as in old versions of Express, the new method is:

res.sendStatus(404);

Express will send a very basic 404 response with "Not Found" text:

HTTP/1.1 404 Not Found
X-Powered-By: Express
Vary: Origin
Content-Type: text/plain; charset=utf-8
Content-Length: 9
ETag: W/"9-nR6tc+Z4+i9RpwqTOwvwFw"
Date: Fri, 23 Oct 2015 20:08:19 GMT
Connection: keep-alive

Not Found

Solution 3 - Javascript

You don't have to simulate it. The second argument to res.send I believe is the status code. Just pass 404 to that argument.

Let me clarify that: Per the documentation on expressjs.org it seems as though any number passed to res.send() will be interpreted as the status code. So technically you could get away with:

res.send(404);

Edit: My bad, I meant res instead of req. It should be called on the response

Edit: As of Express 4, the send(status) method has been deprecated. If you're using Express 4 or later, use: res.sendStatus(404) instead. (Thanks @badcc for the tip in the comments)

Solution 4 - Javascript

According to the site I'll post below, it's all how you set up your server. One example they show is this:

var http = require("http");
var url = require("url");

function start(route, handle) {
  function onRequest(request, response) {
    var pathname = url.parse(request.url).pathname;
    console.log("Request for " + pathname + " received.");

    route(handle, pathname, response);
  }

  http.createServer(onRequest).listen(8888);
  console.log("Server has started.");
}

exports.start = start;

and their route function:

function route(handle, pathname, response) {
  console.log("About to route a request for " + pathname);
  if (typeof handle[pathname] === 'function') {
    handle[pathname](response);
  } else {
    console.log("No request handler found for " + pathname);
    response.writeHead(404, {"Content-Type": "text/plain"});
    response.write("404 Not found");
    response.end();
  }
}

exports.route = route;

This is one way. http://www.nodebeginner.org/

From another site, they create a page and then load it. This might be more of what you're looking for.

fs.readFile('www/404.html', function(error2, data) {
            response.writeHead(404, {'content-type': 'text/html'});
            response.end(data);
        });

http://blog.poweredbyalt.net/?p=81

Solution 5 - Javascript

From the Express site, define a NotFound exception and throw it whenever you want to have a 404 page OR redirect to /404 in the below case:

function NotFound(msg){
  this.name = 'NotFound';
  Error.call(this, msg);
  Error.captureStackTrace(this, arguments.callee);
}

NotFound.prototype.__proto__ = Error.prototype;

app.get('/404', function(req, res){
  throw new NotFound;
});

app.get('/500', function(req, res){
  throw new Error('keyboard cat!');
});

Solution 6 - Javascript

IMO the nicest way is to use the next() function:

router.get('/', function(req, res, next) {
	var err = new Error('Not found');
	err.status = 404;
	return next(err);
}

Then the error is handled by your error handler and you can style the error nicely using HTML.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionRandomblueView Question on Stackoverflow
Solution 1 - JavascriptDrew NoakesView Answer on Stackoverflow
Solution 2 - JavascriptBradView Answer on Stackoverflow
Solution 3 - JavascriptrossipediaView Answer on Stackoverflow
Solution 4 - JavascriptcraniumonemptyView Answer on Stackoverflow
Solution 5 - JavascriptalessioalexView Answer on Stackoverflow
Solution 6 - JavascriptSamView Answer on Stackoverflow