php code problem

[kg]

Hi

I'm using codeigniter for my php project i' already have a drop down list displayed in one page of the web browser which had dynamic values from my database, as soon as you click on the submit button it takes you to another page which has to display the cities from my database and that is not happening at the moment, this code used to work nicely when i had the cities displayed in a single page as  the drop menu but doesn't anymore since i need the cities displayed on another page ,any help will be highly appreciated.

 

<?

function writeCities($id)

{

$con = mysql_connect("localhost","root","");

if (!$con) die('Could not connect: ' . mysql_error());

  mysql_select_db("msansi", $con);

$query  = "SELECT cities FROM provinces WHERE id =";

$query .= $id;

$result = mysql_query($query);

 

$row = mysql_fetch_array($result);

echo $row[0];

}

 

function onChangeDropBox()

 

{

var selected =0;

selected = document.myform.province.value;

 

var t = ["<? writeCities(9);?>";]

document.myform.submit.value = t;

 

}

 

 

[cyberRobot]

I think the issue is that you're trying to call a PHP function from within JavaScript. Unfortunately, you can't do that on the client side. If you want PHP to process the drop-down selection, you need to submit the form back to the server where PHP is processed.

 

What are you trying to accomplish?

[cyberRobot]

Actually, I'm a bit confused. Which page is the above code for? Is it the code for the second page; after the initial form is submitted? We may need to see a little more of the code.

 

 

Also, please use the code tags ([ code ][ /code ]) when posting code. You'll need to remove the spaces from the code tag. I just added them to prevent the forum from interpreting them. Or you could just click the button that looks like a pound/hash sign (#) when composing a message.