Update data from comboBox using PHP - php

Question

I need your help..
In my web site, I let the employee to chose the department name that he wants from comboBox using this code
$dept_id = $_SESSION['dept_id'];
$from= "SELECT d.dept_id, d.name FROM department d LEFT JOIN employee e ON d.dept_id = e.dept_id WHERE emp_id = '$emp_id' ";
$result_form = mysql_query($from);
//$row1 = mysql_fetch_array($result_form); // get 1st result row
$dept_from = mysql_fetch_assoc($result_form);
$dept_name = $dept_from['name'];
//$row=mysql_fetch_array($result_form);
//echo $row1['name'];
$date1=date("Y/m/d");
$dept_id = $_SESSION['dept_id'];
$query = "INSERT INTO request (`date`, `description`, `from`, `emp_id`, `to`)
VALUES
('$date1','$_POST[description]','$dept_name','$emp_id','$_POST[to]')";
it works correctly
Then the administrator choose the request by employee ID from comboBox when he press the accept button the data should updated
and this is its code
if(array_key_exists('accept', $_POST)) {
$conn = mysql_connect("localhost","root","");
mysql_select_db("employee_transfare", $conn);
$emp_id=$_POST['emp_id'];
$dept_id = $_SESSION['dept_id'];
$query="UPDATE employee SET dept_id='$dept_id' WHERE emp_id= $emp_id ";
$n=mysql_query($query, $conn);
if($n==0)
echo "<h2>details already updated </h2>";
else
echo "<h2>details successfully updated</h2>";
mysql_close($conn);
}
else {
It always gives details already updated ...
How I can do it?

Answer 1

That comparison with 0 seems a bit imprecise. The following is from the documentation on mysql_query:
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.
For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.
Instead of if ($n==0), perhaps try if ($n===FALSE) or alternately if (! $n).