Select dropdown value to get a populated data from database mysql - php

Question

I have a populated dropdown list from mysql on one page.
What code should i write in php so that on selection of dropdown list's value, a new page is opened and data is fetched from mysql, provided i have required fields in database.
<form name="form1">
<select>
<?php
$con=mysql_connect("localhost","FreeUser","123456");
if(!$con)
{
die('Connection failed' . mysql_error());
}
mysql_select_db("human_resource", $con);
$sql1=mysql_query("SELECT eid_emp,name_emp FROM employee_details");
while ($data=mysql_fetch_assoc($sql1))
{
?>
<option name="drop1" checked value ="<?php echo $data['eid_emp'] ?>" >
<?php echo $data['name_emp']; ?>
</option>
<?php
}
mysql_close($con);
?>
</select>
</form>

Answer 1

The answer is there is no PHP code to do what you're asking.
You'll need two things: 1). Some javascript to hook into your form and 2). a way to render the resulting query.
here is some javascript to hook into the form, assuming jquery for brevity (although you should just give the select an ID or a class, which would make the selector less obnoxious):
$('form[name="form1"] select').change(function(){
location.href='renderer.php?dataKey=' + $(this).val();
});
From there, you'll navigate to renderer.php along with $_GET value for dataKey. render it as your will. If you want to open a new window, use a window.open call instead of setting location.href.

Answer 2

has nothing to do with php or mysql ... you could add an "onchange" handler to your < select > element ...
<select onchange="javascript:someFunctionCallHere()">
a call to your forms submit() method should be what you want...

Answer 3

Not a lot of information, but you could do something like this:
<form name="form1" method="POST" action="page.php">
<select onchange="form1.submit()">
Then in the head of your page
<?php
if(count($_POST))
{
// do stuff
header( 'Location: http://somesite.com' ) ;