Change dropdown list (options) values from database with ajax and php

I’m going to show you a example in php and ajax to change the values of the dropdown’s options without refreshing the page. The values (options) of the dropdown are fetched from the database and the certain portion of the web pages is only refreshed without need to refresh the whole page.

Let’s start with creating the two tables country and city and insert some data

CREATE TABLE country
(
   id tinyint(4) NOT NULL auto_increment,
   country varchar(20) NOT NULL default '',
   PRIMARY KEY  (id)
) TYPE=MyISAM;CREATE TABLE city
(
  id tinyint(4) NOT NULL auto_increment,
  city varchar(50) default NULL,
  countryid tinyint(4) default NULL,
  PRIMARY KEY  (id)
) TYPE=MyISAM;

Now let’s look at the html code, let’s look at the code of the form and its elements

<form method="post" action="" name="form1">
Country : <select name="country" nChange="getCity('findcity.php?country='+this.value)">
 <option value="">Select Country</option>
 <option value="1">USA</option>
 <option value="2">Canada</option>
     </select>
<br />City : <div id="citydiv">
 <select name="select">
 <option>Select City</option>
     </select>
 </div>
</form>

As you can see that when the dropdown named “country” is changed the “getCity” function is called. Look at the other dropdown carefully, it is inside the division called “citydiv”.

Now let’s look at the javascript function called “getCity”

function getCity(strURL)
{         
 var req = getXMLHTTP(); // fuction to get xmlhttp object
 if (req)
 {
  req.onreadystatechange = function()
 {
  if (req.readyState == 4) { //data is retrieved from server
   if (req.status == 200) { // which reprents ok status                    
     document.getElementById('citydiv').innerHTML=req.responseText;
  }
  else
  { 
     alert("There was a problem while using XMLHTTP:\n");
  }
  }            
  }        
req.open("GET", strURL, true); //open url using get method
req.send(null);
 }
}

now we’ve to create the file called findcity.php and put the following PHP code

<? $country=$_GET['country'];
$link = mysql_connect('localhost', 'root', ''); /change the configuration if required
if (!$link) {
  die('Could not connect: ' . mysql_error());
}
mysql_select_db('db_ajax'); //change this if required
$query="select city from city where countryid=$country";
$result=mysql_query($query);?>
<select name="city">
<option>Select City</option>
<? while($row=mysql_fetch_array($result)) { ?>
   <option value><?=$row['city']?></option>
<? } ?>
</select>

Thats all, whenever you change the dropdown of country the values of the city dropdown is automaticalled changed without refreshing the page.
If you want to download full source code, Click here to download.

Security Note : If you want to use this code in your project then there is a security flaw in the PHP code, please use $country=intval($_GET['country']); in the php code instead of $country=$_GET['country']; in the findcity.php to prevent your site from sql injection attack.