Adding more parameters in ajax - php

Question

I am using AJAX function. I am passing 3 variables to the next page using AJAX. When I add the 4th variable the function doesn't get called.
Code:
<script language="javascript" type="text/javascript">
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
var count = document.getElementById('count').value;
var type = document.getElementById('type').value;
var sem = document.getElementById('sem').value;
var rid = document.getElementById('room').value;
ajaxRequest.open("GET", "add_requi_ajax.php?count=" + count+"&type="+ type+"&sem="+sem+"&rid="+rid, true);
ajaxRequest.send(null);
};
</script>

Answer 1

Your code is syntactically and logically correct, which means that the problem is likely one of your input IDs is wrong (typo? Should room be rid?), or you call the function before the inputs are rendered on the page (use window.onload).
Verify each of your input IDs. If they all look correct, then comment them out and hard code the values to rule out your inputs as a problem. Watch the error console for any error messages. If an uncaught error is encountered, it can appear that the function isn't being called.

Answer 2

You probably need to urlencode your values.