[SOLVED] Php/Ajax/Mysql problem

jv · August 20, 2008

Hi guys, im pretty new to ajax and thought i was pretty okish at using mysql and php. Im working on a webpage that allows you to input a persons name and print out competition results associated with that name. I used the following tutorial to learn ajax and my code is near identical. http://www.tizag.com/ajaxTutorial/ajax-mysql-database.php

Index.html:

<html>
<body>

<script language="javascript" type="text/javascript">
<!--
function ajaxFunction(){
var ajaxRequest;
try{
	ajaxRequest = new XMLHttpRequest();
} catch (e){
	try{
		ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
	} catch (e) {
		try{
			ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
		} catch (e){
			alert("Your browser broke!");
			return false;
		}
	}
}
ajaxRequest.onreadystatechange = function(){
	if(ajaxRequest.readyState == 4){
			var ajaxDisplay = document.getElementById('ajaxDiv');
					ajaxDisplay.innerHTML = ajaxRequest.responseText;
				}

}
var age = document.getElementById('name').value;
var queryString = "?name=" + name;
ajaxRequest.open("GET", "getresults.php" + queryString, true);
ajaxRequest.send(null);
}

//-->
</script>



<form name='myForm'>
Competitor Name: <input type='text' id='name' /> <br />
</select>
<input type='button' onclick='ajaxFunction()' value='Query MySQL' />
<div id='ajaxDiv'>Your result will display here</div>
</form>
</body>
</html>

getresults.php:

<?php
$dbhost = '';
$dbuser = '';
$dbpass = '';
$dbname = '';

mysql_connect($dbhost, $dbuser, $dbpass);

mysql_select_db($dbname) or die(mysql_error());

$name = $_GET['name'];

$name = mysql_real_escape_string($name);

$query = "SELECT * FROM tablename WHERE Name = '$name'";

$qry_result = mysql_query($query) or die(mysql_error());

$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>Club</th>";
$display_string .= "<th>FTot</th>";
$display_string .= "<th>Totals</th>";
$display_string .= "</tr>";

while($row = mysql_fetch_array($qry_result)){
$display_string .= "<tr>";
$display_string .= "<td>$row[Name]</td>";
$display_string .= "<td>$row[Club]</td>";
$display_string .= "<td>$row[Ftot]</td>";
$display_string .= "<td>$row[Totals]</td>";
$display_string .= "</tr>";

}
echo "Query: " . $query . "<br />";
$display_string .= "</table>";
echo $display_string;
?>

as you can see the code prints out the query on the page before showing the results. but the query printed is

Query: SELECT * FROM Comp2 WHERE Name = ''

as if it is cut short.

Can anyone point me in the right direction?

Thanks

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[SOLVED] Php/Ajax/Mysql problem

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