I'm trying to create a program that when you select a state from the drop down menu, it will display the list of cities for that state in another drop down menu that you can select from. After you choose your city and state, you type in an address, hit submit, and it will display the full address on a new php file.
For example, if you select New Jersey from the drop down menu, the drop down menu to the left of it should display three cities: Newark, Bloomfield and Edison.
My issue at the moment is I can get the states displayed, but when the state is selected, it is not giving me the list of options for that city in the second drop down menu. Any help is appreciated, thanks!
You can view this behavior at this link
select.php
<head>
<link rel="stylesheet" type="text/css" href="select_style.css">
<script type="text/javascript" src="js/jquery.js"></script>
<!DOCTYPE html>
<form action = "display.php">
<script type="text/javascript">
function fetch_select(val)
{
$.ajax({
type: 'get',
url: 'fetch.php',
data: {
get_option:val
},
success: function (response) {
document.getElementById("new_select").innerHTML=response;
}
});
}
</script>
</head>
<body>
<p id="heading">Address Generator</p>
<center>
<div id="select_box">
<select onchange="fetch_select(this.value);">
<option>Select state</option>
<?php
include ( "accounts.php" ) ;
( $dbh = mysql_connect ( $hostname, $username, $password ) )
or die ( "Unable to connect to MySQL database" );
print "Connected to MySQL<br>";
mysql_select_db( $project );
$select=mysql_query("select state from zipcodes group by state");
while($row=mysql_fetch_array($select))
{
echo "<option value='".$row['state']."'>".$row['state']."</option>";
}
?>
</select>
<select id="new_select">
</select>
<div id='2'> </div>
<br><br>
<input type = text name="address">Address
<br><br>
<input type = submit>
</form>
fetch.php
<?php
include(accounts.php);
if(isset($_POST['get_option']))
{
( $dbh = mysql_connect ( $hostname, $username, $password ) )
or die ( "Unable to connect to MySQL database" );
print "Connected to MySQL<br>";
mysql_select_db( $project );
$state = $_GET['get_option'];
$find=mysql_query("select city from zipcodes where state='$state'");
while($row=mysql_fetch_array($find))
{
echo "<option>".$row['city']."</option>";
}
exit;
}
?>
