I have got 2 php scripts with javascript as shown below.
I got the dynamic drop down list to work but i do not know how to submit those values into another table in database. Basically the scenario is, I want the user to select values from the drop down list and when he clicks submit button, the values will be stored in another table (called booking) in the mysql database.
Any ideas?
The 2 current tables i got that goes with the scripts below are:
Countries and Cities
These are the scripts i got
index.php page
Code: Select all
<?php
$conn = mysql_connect("localhost", "root", "");
$db = mysql_select_db("test", $conn);
$sql_country = "SELECT * FROM country";
$result_country = mysql_query($sql_country);
?>
<!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>
<title>Country List</title>
</head>
<body>
<?php
echo "<select name='country' onChange='get_cities(this.value)'>";
while($row_country = mysql_fetch_array($result_country))
{
echo "<option value='".$row_country['id']."'>".$row_country['country']."</option>";
}
echo "</select>";
echo "<div id='cityLayer'><select name='city' id='city'></select></div>";
?>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
function get_cities($country_id){
$.ajax({
url : "city.php?country_id="+$country_id,
cache : false,
beforeSend : function (){
//Show a message
},
complete : function($response, $status){
if ($status != "error" && $status != "timeout") {
$('#cityLayer').html($response.responseText);
}
},
error : function ($responseObj){
alert("Something went wrong while processing your request.\n\nError => "
+ $responseObj.responseText);
}
});
}
</script>
</body>
</html>Code: Select all
<?php
$conn = mysql_connect("localhost", "root", "");
$db = mysql_select_db("test", $conn);
$country_id = $_REQUEST['country_id'];
$sql_city = "SELECT * FROM cities WHERE country_id = '".$country_id."'";
$result_city = mysql_query($sql_city);
echo "<select name='city'>";
while($row_city = mysql_fetch_array($result_city))
{
echo "<option value='".$row_city['id']."'>".$row_city['city']."</option>";
}
echo "</select>";
?>
