Hi I am a Ajax noobie and need a little help with a simple ajax mysql problem

I have a php page witha ajax function that calls a php page and in the php page I run a mysql select

I then pass the result back to the page that the ajax is on and put the result into a div

all good.

my question is, how do I run an if statement on the passed back string

ajax page

<html>
<body>
<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 c_id = document.getElementById('c_id').value;	
	var queryString = "?c_id=" + c_id;
	ajaxRequest.open("GET", "serverTime.php" + queryString, true);
	ajaxRequest.send(null); 
}

//-->
</script>

<form name='myForm'>
      c_id: <input type='text' id='c_id'  /> <br />
     <input type='button' onclick='ajaxFunction();' value='Query MySQL' />
</form>

<div id='ajaxDiv'>
     <? echo $url; ?>
</div>


</body>
</html>

and php page

<?php
require("connect.php");
$c_id = $_GET['c_id'];

$query = mysql_query("SELECT * FROM chat_pms WHERE target = '$c_id'");

while($row=mysql_fetch_array($query))
	{	
		$display_string = $row['name'];			
	}		

echo $display_string;
?>

I have tried a php if but obviously that wont work as php only runs on page load.

I have tried a javascript if and can't get that to work either as my javascript is not the best

any help greatly appreciated

if(ajaxRequest.responseText=="what ever your looking to return")
Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.