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?

Marcos Casagrande · Accepted Answer · 2019-03-09 22:10:17Z

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;

edited Mar 9, 2019 at 22:10

answered Mar 9, 2019 at 22:01

Marcos Casagrande

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11 Comments

Tarneem Over a year ago

I am using bodyParser, but how to access each one of these variable from the server? as req.body is only picking the first variable in xhttp.send(x,y);

Tarneem Over a year ago

req.body.x; is causing error I dont know why, I have to use req.body only. using this: const { x, y } = req.body; is not getting me right values

Marcos Casagrande Over a year ago

Show your updated code, my code example works fine, you're doing something wrong. Post server & client code.

Tarneem Over a year ago

client side: xhttp.send(JSON.stringify({ x: x, y:y}));

Tarneem Over a year ago

server side: const x = req.body.x; const y = req.body.y;

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