I am absolutely newbie to Ajax and I need to have this fixed by tomorrow as I have a project to complete so anyone who could help me, I would be really greatful!
This is what I have cooked up so far
index.php
<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 deviceid = document.getElementById('deviceid').value;
var queryString = "?deviceid=" + deviceid;
ajaxRequest.open("GET", "data.php" + queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
<form name='myForm' method="post" action="test.php">
Device ID: <input type='text' id='deviceid' /> <br />
<input type='button' onclick='ajaxFunction()' value='Query MySQL' />
<div id='ajaxDiv'>
</div>
<input name="Submit" type="submit" value="submit" />
</form>
</body>
</html>data.php
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "ajax";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
// Retrieve data from Query String
$deviceid = $_GET['deviceid'];
// Escape User Input to help prevent SQL Injection
$deviceid = mysql_real_escape_string($deviceid);
//build query
$query = "SELECT * FROM pmc_devicedata_100902 WHERE DeviceID = '$deviceid'";
$qry_result = mysql_query($query) or die(mysql_error());
//Build Result String
// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
$aux = $row['AuxBarcode'];
}
$display_string .= "Aux:<input type='text' name='aux' value='$aux' disabled />";
echo $display_string;
?>test.php
<?php
$aux = $_POST['aux'];
echo "AUX: $aux";
?>test.php does not print any results. I think AJAX Is just displaying the results but not passing the value to the PHP script (test.php). This is a very basic script which I plan to improve it once I get this one to work. Any help provided would be appreciated