How to retrieve data from xhttp.send() in node.js express

Question

so I have this line in front page:

 var x = "Hi"; var y = 123; xhttp.open("POST", "/toNodeServer", true); xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); xhttp.send(x, y); 

and in the server I have the following:

 outer.post('/toNodeServer', function(req, res, next){ var x = req.body console.log(x); 

so the result is it does not send y value, I got this from the terminal: {'Hi': ''}

can anyone please explain what is going on and how to send these two variables?

Answer 1

XMLHttpRequest.send(body) only takes 1 parameter.

You need to post a JSON or send an url-encoded string (or any other serialized string)

JSON

xhttp.setRequestHeader("Content-type", "application/json");
xhttp.send(JSON.stringify({ x: x, y: y}));

x-www-form-urlencoded

xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send(`x=${x}&y=${y}`);

If you're sending JSON , don't forget to add:

app.use(bodyParser.json());
app.use(bodyParser.urlencoded());

Then you will be able to access it this way:

const x = req.body.x;
const y = req.body.y;

Or using destructuring:

const { x, y } = req.body;