Hi, I have some problem in ajax like i am not able to retrieve data through Ajax. Please look at the attachment pic to better understand the scenario. I am trying to get all the values from a database which have 4 states and 3 cities in each state and third 2 or three market in each citi. Here i am getting issue for the first time when the page is first time loaded. I have called a javascript function on first page which gets the value for state and cities is loaded but the market menu remains the blank. I want to get the value in market according the whatever city is selected on the city tab. Hope you can understand what i am looking for. I have tried to find out the problem but all in vain. Here i am attaching all the php code below and hop i need not to upload database details but if you require then i will add too. Thank in advance. Here is the main designing file code: <!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=utf-8" /><title>Untitled Document</title><script type="text/javascript">function Action(link,md){ var obj,url; url="dcode.php?mode="+md+"&id="+link+"&r="+Math.random(); try { obj=new XMLHttpRequest(); } catch(e) { try { obj=new ActivexObject("Microsoft.XMLHTTP"); } catch(e) { alert(e); } } obj.open("GET",url,true); obj.send(null); obj.onreadystatechange=function() { if(obj.readyState==4) { var res=obj.responseText; if(md=='city') { document.getElementById("market").innerHTML=res; } else { document.getElementById("city").innerHTML=res; } //document.getElementById("city").innerHTML=res; } } }</script> <? ?> </head> <body> <form action="" method="get"><table width="200" border="1" align="center"> <tr> <th scope="row"><label> <select name="state" id="state" style="width:150px;" onchange="Action(this.value,'state')"> <? mysql_connect("localhost","root","") or die("check server"); mysql_select_db("students")or die("check database"); $str="select sno,sname from state"; $rs=mysql_query($str); while(list($scode,$stname)=mysql_fetch_array($rs)) { echo "<option value='$scode'>$stname</option>"; } ?> </select> </label></th> </tr> <tr> <th scope="row"><label> <select name="city" id="city" style="width:150px;" onchange="Action(this.value,'city')"> </select> </label></th> </tr> <tr> <th scope="row"><select name="market" id="market" style="width:150px;"> </select></th> </tr> </table></form> <script type="text/javascript">Action(101,'state'); </script></body></html> PHP: Here is the Code file code: <? mysql_connect("localhost","root","") or die("check server");mysql_select_db("students")or die("check database"); if($_REQUEST["mode"]==state) { $id=$_REQUEST["id"]; $str="SELECT citycode,cityname FROM city WHERE statecode IN ( SELECT statecode FROM city JOIN state ON city.statecode = $id)"; $rs=mysql_query($str); while(list($citycode,$cityname)=mysql_fetch_array($rs)) { echo "<option value='$citycode'>$cityname</option>"; } } if($_REQUEST["mode"]==city) { $cid=$_REQUEST["id"]; $str="SELECT mcode,mname FROM market WHERE citicode IN ( SELECT citicode FROM market JOIN city ON market.citicode = $cid)"; $rs=mysql_query($str); while(list($mcode,$mname)=mysql_fetch_array($rs)) { echo "<option value='$mcode'>$mname</option>"; } } ?> PHP:
Could you make your code more readable? And also do you have firebug extension installed? It has the ability to analyze all the ajax requests and responses.
Well the code was readable but digital-point editor made it look like this. May be copying and pasting these code on notepad ++ or other editor may help you. I haven't tried the firebug extension for this issue will try it.