So you have the onchange in there and that's a start. The onchange is referencing a JavaScript function that you don't show. There are a couple quick ways to approach this:
- Post the form to itself (as you have chosen) or
- use ajax (possibly via jQuery for quickness).
(both of these examples don't address how you are accessing the database)
1)
Using your run function:
<script type="text/javascript">
function run(){
document.getElementById('form1').submit()
}
</script>
Additional PHP:
<?php
if (isset($_POST['department']) && isset($_POST['type_hire']))
{
$value_of_department_list = $_POST['department'];
$value_of_type_hire = $_POST['type_hire'];
mysql_connect("localhost","root","");
mysql_select_db("employees_hired");
mysql_query("SELECT name FROM usuario WHERE (department ='". $value_of_department_list ."') AND (contrasena ='". $value_of_type_hire."')");
while($row_list=mysql_fetch_assoc($list))
{
echo "<option value="{$row_list['name']}">{$row_list['name']}</option>";
}
}
else
{
echo "<option>Please choose a department and a type of hire</option>";
}
?>
2)
<script type="text/javascript">
function run(){
$.post('get_employees.php',$('form1').serialize(),function(data){
var html = '';
$.each(data.employees,function(k,emp){
$('select[name="employees"]').append($('<option>', {
value: emp.name,
text: emp.name
}));
.html(html);
},"json");
}
</script>
And get_employees.php would contain something like:
<?php
if (isset($_POST['department']) && isset($_POST['type_hire']))
{
$value_of_department_list = $_POST['department'];
$value_of_type_hire = $_POST['type_hire'];
$return = array();
mysql_connect("localhost","root","");
mysql_select_db("employees_hired");
mysql_query("SELECT name FROM usuario WHERE (department ='". $value_of_department_list ."') AND (contrasena ='". $value_of_type_hire."')");
while($row_list=mysql_fetch_assoc($list))
{
$return[]['name'] = $row_list['name'];
}
echo json_encode($return);
}
?>
Note, these are just quickly written examples. A lot more could/should be done here.
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