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

<script language="javascript" type="text/javascript">
//Browser Support Code
function ajaxFunction(){
	var ajaxRequest;  // The variable that makes Ajax possible!
		// Opera 8.0+, Firefox, Safari
		ajaxRequest = new XMLHttpRequest();
	} catch (e){
		// Internet Explorer Browsers
			ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
		} catch (e) {
				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);


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

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


and php page

$c_id = $_GET['c_id'];

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

		$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

7 Years
Discussion Span
Last Post by wrivera
This topic has been dead for over six months. Start a new discussion instead.
Have something to contribute to this discussion? Please be thoughtful, detailed and courteous, and be sure to adhere to our posting rules.