Answers for "request.open get url"

9

javascript send post request

let data = {element: "barium"};

fetch("/post/data/here", {
  method: "POST", 
  body: JSON.stringify(data)
}).then(res => {
  console.log("Request complete! response:", res);
});


// If you are as lazy as me (or just prefer a shortcut/helper):

window.post = function(url, data) {
  return fetch(url, {method: "POST", body: JSON.stringify(data)});
}

// ...

post("post/data/here", {element: "osmium"});
Posted by: Guest on March-04-2020
3

javascript http request

// Use these functions:

function _GET_REQUEST(url, response) {
  var xhttp;
  if (window.XMLHttpRequest) {
    xhttp = new XMLHttpRequest();
  } else {
    xhttp = new ActiveXObject("Microsoft.XMLHTTP");
  }

  xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
      response(this.responseText);
    }
  };

  xhttp.open("GET", url, true);
  xhttp.send();
}

function _POST_REQUEST(url, params, response) {
  var xhttp;
  if (window.XMLHttpRequest) {
    xhttp = new XMLHttpRequest();
  } else {
    xhttp = new ActiveXObject("Microsoft.XMLHTTP");
  }

  xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
      response(this.responseText);
    }
  };

  xhttp.open("POST", url, true);
  xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
  xhttp.send(params);
}


// Use like:
_GET_REQUEST('http://url.com', (response) => {
	// Do something with variable response
  	console.log(response);
});
_POST_REQUEST('http://url.com', 'parameter=sometext', (response) => {
	// Do something with variable response
  	console.log(response);
});
Posted by: Guest on October-18-2020

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