Answers for "post request with xhr"

xmlhttp js post request

var xhr = new XMLHttpRequest();
xhr.open("POST", '/url', true);
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");

xhr.onreadystatechange = function() { // Call a function when the state changes.
    if (this.readyState === XMLHttpRequest.DONE && this.status === 200) {
        // Request finished. Do processing here.
    }
}
xhr.send("name=Hello&id=world");

javascript send post data with ajax

function makeRequest (method, url, data) {
  return new Promise(function (resolve, reject) {
    var xhr = new XMLHttpRequest();
    xhr.open(method, url);
    xhr.onload = function () {
      if (this.status >= 200 && this.status < 300) {
        resolve(xhr.response);
      } else {
        reject({
          status: this.status,
          statusText: xhr.statusText
        });
      }
    };
    xhr.onerror = function () {
      reject({
        status: this.status,
        statusText: xhr.statusText
      });
    };
    if(method=="POST" && data){
        xhr.send(data);
    }else{
        xhr.send();
    }
  });
}

//POST example
var data={"person":"john","balance":1.23};
makeRequest('POST', "https://www.codegrepper.com/endpoint.php?param1=yoyoma",data).then(function(data){
              var results=JSON.parse(data);
});

ajax open a request

<script>
function loadDoc() {
  var xhttp = new XMLHttpRequest();
  
  //looking for a change or state , like a request or get.
  xhttp.onreadystatechange = function() {
     
     //if equal to 4 means that its ready.
    // if equal to 200 indicates that the request has succeeded.
    if (this.readyState == 4 && this.status == 200) {
      document.getElementById("demo").innerHTML = this.responseText;
    }
  };
  
  //GET method for gettin the data // the file you are requesting
  xhttp.open("GET", "TheFileYouWant.html", true);
  
  //sending the request
  xhttp.send();
}