Answers for "ajax open url"

2

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();
}
Posted by: Guest on January-18-2020

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