rajmohan Posted August 9, 2007 Share Posted August 9, 2007 here is the file called HHAajaxResponse.php <?php $dbhost = "localhost"; $dbuser = "root"; $dbpass = ""; $dbname = "testing"; //Connect to MySQL Server mysql_connect($dbhost, $dbuser, $dbpass); //Select Database mysql_select_db($dbname) or die(mysql_error()); // Retrieve data from Query String $subregion = $_GET['subregion']; // Escape User Input to help prevent SQL Injection $subregion = mysql_real_escape_string($subregion); //build query $query = "SELECT * FROM suburbs WHERE city_id = '$subregion'"; //Execute query $qry_result = mysql_query($query) or die(mysql_error()); // Insert a new row in the table for each person returned while($row = mysql_fetch_array($qry_result)){ $display_string .= "<option value=".$row[suburb_id].">".$row[suburb_name]."</option>"; //$display_string .= "<option value=\"$row[suburb_id]\">$row[suburb_name]</option>"; } echo "$display_string"; ?> and html file as index.php <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>ajaxForm</title> <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('suburbs'); ajaxDisplay.innerHTML = ajaxRequest.responseText; } } var subregion = document.getElementById('subregion').value; var queryString = "?subregion=" + subregion; alert(queryString); ajaxRequest.open("GET", "HHAajaxResponse.php" + queryString, true); ajaxRequest.send(null); alert } function syncbox() { sexlist = document.myForm.sex.value; document.myForm.sexupdate.value = sexlist; } function sync(){ interval = setInterval("syncbox()",1); } //--> </script> </head> <body> <form id="form1" name="form1" method="post" action=""> <label>area <select name="select" id="subregion" onchange="ajaxFunction()"> <option value="0">select</option> <option value="1">durban</option> <option value="2">joburg</option> <option value="3">capetown</option> </select> </label> <label></label> <p> <label>sub area</label> <select name="suburbs" id="suburbs"> </select> </p> </form> </body> </html> Quote Link to comment Share on other sites More sharing options...
rajmohan Posted August 9, 2007 Author Share Posted August 9, 2007 my table CREATE TABLE `suburbs` ( `city_id` int(11) NOT NULL, `suburb_id` int(11) NOT NULL, `suburb_name` varchar(30) NOT NULL ) ENGINE=InnoDB DEFAULT CHARSET=latin1; # # Dumping data for table `suburbs` # INSERT INTO `suburbs` VALUES (1, 1, 'one'); INSERT INTO `suburbs` VALUES (1, 2, 'two'); INSERT INTO `suburbs` VALUES (2, 1, 'ha'); INSERT INTO `suburbs` VALUES (2, 2, 'va'); INSERT INTO `suburbs` VALUES (3, 1, 'asd'); INSERT INTO `suburbs` VALUES (3, 2, 'asdf'); Quote Link to comment Share on other sites More sharing options...
JJohnsenDK Posted August 9, 2007 Share Posted August 9, 2007 do tell more... i have no clue what your problem is? and where the problem is? Quote Link to comment Share on other sites More sharing options...
rajmohan Posted August 9, 2007 Author Share Posted August 9, 2007 if I put the recieved data from the PHP file I recieve all the info but as soon as I put it into a drop-down list the list does not populate. it just shows up blank. Quote Link to comment Share on other sites More sharing options...
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