Member Avatar for narasimha9

828915c19b3102a17ba1a5040e2a76ee828915c19b3102a17ba1a5040e2a76ee828915c19b3102a17ba1a5040e2a76ee I want to retrieve values from database to seleced box.I have three files ajax.js,getcitylist.php,securepage.php.In second select box values are not coming from data base.

    <li>
        <label for="State">State</label>
        <select id="state"  name="state" onChange="display(this.value)">
        <option value="" selected="selected">-- Select state --</option>
        <?php include("getstatelist.php");?>
        </select>
    </li>
    <li>
        <label for="City">City</label>
        <div id="city">
        <select id="city"  name="city" onChange="display(this.value)">
        <option value="" selected="selected">-- Select city --</option>
        <?php include("getcitylist.php");?>
        </select>
        </div>
    </li>
Edited by narasimha9 because: attached image of citylist
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Member Avatar for narasimha9

ajax.js

var XMLHttpRequestObject=false;
function display(state_id)
{
if(window.XMLHttpRequest)
{
XMLHttpRequestObject=new XMLHttpRequest();
}
else if(window.ActiveXObject)
{
XMLHttpRequestObject=new ActiveXObject("Microsoft.XMLHTTP");
} 
XMLHttpRequestObject.onreadystatechange=function()
{
if (XMLHttpRequestObject.readyState==4 && XMLHttpRequestObject.status==200)
{
document.getElementById("city").innerHTML=XMLHttpRequestObject.responseText;
}
}
XMLHttpRequestObject.open("GET","getcitylist.php?state_id,city_id="+state_id,+city_id,true);
XMLHttpRequestObject.send();
}
Member Avatar for narasimha9 
<?php
include("config.php");
/*$con=mysql_connect('localhost','root','') or die('Mysql not connected');
mysql_select_db('location',$con) or die('DataBase not connected');*/

$state_id=$_REQUEST['state_id'];
$city_id=$_REQUEST['city_id'];

//$query="select * from tbl_city where state_id='$state_id'";
$query="select * from tbl_city where state_id='$state_id' and  city_id='$city_id'";

?>

 <select id="city"  name="city" onChange="display(this.value)">
 <option value="" selected="selected">-- Select city --</option>
<?php

$query_result=mysql_query($query)or mysql_error();
while($row=mysql_fetch_array($query_result))
{
?>
<option value="<?php echo $row['id']; ?>"><?php echo $row['city_name']; ?></option>
<?php
}
?>
Member Avatar for narasimha9

first file is secure.php second file is ajax.js and third one is getcitylist.php

In getcitylist.php I feel there is wrong with query. what could be the query?

Member Avatar for Lsmjudoka

Your AJAX call is sending the request like this:

19. XMLHttpRequestObject.open("GET","getcitylist.php?state_id,city_id="+state_id,+city_id,true);

It looks like you're trying to separate the variables in the URL string with a comma and then add the values after, but you need to do them individually. When you are passing paramaters via the GET method, they should be attached to the URL string with ? and &, for example:

xmlhttp.open("GET","demo_get2.asp?fname=Henry&lname=Ford",true);

Try rewriting the URL section of line 19 to fit the example above.

Edit: Also I mentioned this in another thread you posted but in this part:

6.  $state_id=$_REQUEST['state_id'];
7.  $city_id=$_REQUEST['city_id'];
8.
9.  //$query="select * from tbl_city where state_id='$state_id'";
10. $query="select * from tbl_city where state_id='$state_id' and  city_id='$city_id'";

You must sanitize any data received from GET, POST, etc before putting it into an SQL query or else your site will be incredibly vulnerable to hacking via SQL injection.

Edited by Lsmjudoka
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