hello, I managed to populate data from database into selectbox using ajax but i could not managed to have the same data in the text area. Can anybody help me please. <html> <head> <title>Roshan's Triple Ajax dropdown code</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <script language="javascript" type="text/javascript"> // Roshan's Ajax dropdown code with php // This notice must stay intact for legal use // Copyright reserved to Roshan Bhattarai - nepaliboy007@yahoo.com // If you have any problem contact me at http://roshanbh.com.np function getXMLHTTP() { //fuction to return the xml http object var xmlhttp=false; try{ xmlhttp=new XMLHttpRequest(); } catch(e) { try{ xmlhttp= new ActiveXObject("Microsoft.XMLHTTP"); } catch(e){ try{ xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); } catch(e1){ xmlhttp=false; } } } return xmlhttp; } function getState(countryId) { var strURL="findState.php?country="+countryId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) { // only if "OK" if (req.status == 200) { document.getElementById('statediv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } function getCity(countryId,stateId) { var strURL="findCity.php?country="+countryId+"&state="+stateId; var req = getXMLHTTP(); if (req) { req.onreadystatechange = function() { if (req.readyState == 4) { // only if "OK" if (req.status == 200) { document.getElementById('citydiv').innerHTML=req.responseText; } else { alert("There was a problem while using XMLHTTP:\n" + req.statusText); } } } req.open("GET", strURL, true); req.send(null); } } </script> </head> <body> <form method="post" action="" name="form1"> <table width="600" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <td width="150">Country</td> <td width="150"> <? $countryId=intval($_GET['country']); $stateId=intval($_GET['state']); $link = mysql_connect('localhost', 'root', 'emre'); //changet the configuration in required if (!$link) { die('Could not connect: ' . mysql_error()); } mysql_select_db('db_ajax'); $query="SELECT id, country FROM country "; $result=mysql_query($query); ?> <select name="country" onChange="getState(this.value)"> <option>Select City</option> <? while($row=mysql_fetch_array($result)) { ?> <option value="<?=$row['id']?>"><?=$row['country']?></option> <? } ?> </select> </td> <td width="150" align="center">State</td> <td width="150"><div id="statediv"><select name="state" > <option>Select Country First</option> </select></div></td> <td width="150" align="center">City </td> <td width="150"><div id="citydiv"><select name="city"> <option>Select State First</option> </select></div></td> </tr> </table> <table width="600" align="center"> <tr> <td><label> <textarea name="textarea" id="textarea" cols="45" rows="5"></textarea> </label></td> </tr> </table> </form> </body> </html> PHP:
You may want to look into a javascript framework. They make tasks like this really easy. Prototype, Mootools, and Jquery are 3 of the more popular ones. Usually the easiest way is to have the PHP script return JSON data, which javascript can easily parse and properly complete the form or fields.