CODEWITHSUNDEEP
phpajaxhtmlapache
Prasana Ramesh
Prasana Ramesh

Reputation: 97

how to make multiple ajax calls on same html page?

I have the following html page welcoming the user. I intend to make one call to /register once and then to /login once through ajax calls. But when this page renders, clicking the buttons does nothing and the functions showdata() and reg_user() are not even called. What could be the problem? Chrome developer tool says the following

XMLHttpRequest cannot load https: //127.0.0.1/login. No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin 'http: //127.0.0.1' is therefore not allowed access.

But I am not making any Cross domain calls. I'm calling to the same localhost domain after all. Please elucidate what is going on. I'm runnning Apache/Ubuntu/PHP on the server 

<html>
<title> Welcome </title>
<script>
function makeRequestObject(){
   var xmlhttp=false;
   try {
      xmlhttp = new ActiveXObject('Msxml2.XMLHTTP');
   } catch (e) {
      try {
         xmlhttp = new ActiveXObject('Microsoft.XMLHTTP');
      } catch (E) {
         xmlhttp = false;
      }
   }
   if (!xmlhttp && typeof XMLHttpRequest!='undefined') {
      xmlhttp = new XMLHttpRequest();
   }
   return xmlhttp;
}

function showdata()
{
   alert("show data called");
   var xmlhttp=makeRequestObject();
   var user=document.getElementById('user_name').value;
   var pass=document.getElementById('password').value;
   var url = 'https://127.0.0.1/login';
   var params = 'uname='+user+'&'+'pass='+pass;

   xmlhttp.open('POST', url, true);
   xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

   xmlhttp.onreadystatechange=function() {
      if (xmlhttp.readyState==4 && xmlhttp.status == 200) {
         var content = xmlhttp.responseText;
         if( content ){
            document.getElementById('userdiv').innerHTML = content;
         }
      }
   }
   xmlhttp.send(params);
}
function reg_user()
{
   alert("reg_user called");
   var xmlhttp=makeRequestObject();
   var user=document.getElementById('reg_username').value;
   var pass=document.getElementById('reg_password').value;
   var url = 'https://127.0.0.1/register';
   var params = 'uname='+user+'&'+'pass='+pass;

   xmlhttp.open('POST', url, true);
   xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded"$

   xmlhttp.onreadystatechange=function() {
      if (xmlhttp.readyState==4 && xmlhttp.status == 200) {
         var content = xmlhttp.responseText;
         if( content ){
            document.getElementById('regdiv').innerHTML = content;
         }
      }
   }
   xmlhttp.send(params);
}
</script>
<body>
Enter user name : <input type="text" id="user_name" /><br/>
Enter your password : <input type="text" id="password"/><br/>
<input type="button" onclick="showdata()" value="Submit"><br/><br/>
<div id="userdiv">
</div>
<br />
Registration
<br />
Enter user name: <input type="text" id="reg_username" /><br/>
Enter your password : <input type="text" id="reg_password"/><br/>
<input type="button" onclick="reg_user()" value="Submit"><br/><br/>
<div id="regdiv">
</div>
</body>
</html>

Upvotes: 0

Views: 448

Answers (1)

anthumchris
anthumchris

Reputation: 9090

The error you're receiving is caused by CORS validation. Ensure that the server https://127.0.0.1/sets the proper HTTP response headers per the CORS specification.

Upvotes: 1

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