How to capture a drop-down selection and insert into DB?

How can I insert a drop-down value into a DB? I have a form that selects a list of people. And with them i have their unique id showing. So populating isn't he problem, but inserting is... any ideas?

Code:

<?php
$q=$_GET["q"];

$con = mysql_connect('localhost', 'peter', 'abc123');
if (!$con)
 {
  die('Could not connect: ' . mysql_error());
   }

    mysql_select_db("ajax_demo", $con);

    $sql="SELECT * FROM user WHERE id = '".$q."'";

    $result = mysql_query($sql);

     echo "<table border='1'>
   <tr>
  <th>Firstname</th>
   <th>Lastname</th>
   <th>Age</th>
    <th>Hometown</th>
    <th>Job</th>
  </tr>";

      while($row = mysql_f开发者_C百科etch_array($result))
      {
       echo "<tr>";
      echo "<td>" . $row['FirstName'] . "</td>";
     echo "<td>" . $row['LastName'] . "</td>";
     echo "<td>" . $row['Age'] . "</td>";
        echo "<td>" . $row['Hometown'] . "</td>";
    echo "<td>" . $row['Job'] . "</td>";
     echo "</tr>";
     }
      echo "</table>";

         mysql_close($con);
        ?> 

---

<option value="1">John Doe</option>

You get the 1 via post, I don't see where's the problem with inserting that into the database.

Update after adding code to the question:

Please show how you generate the options, this code seems ok. One thing though: you should escape $q before passing it to the query like this: mysql_real_escape_string($q). It returns a string, so you can just concatenate it to the query.
Read about SQL injection attacks.