title: HTML/PHP Survery not passing ID from table correctly
body: <p>I'll try to keep it simple... this is the code I am using to populate a dropdown menu from a database. This populates the dropdown menu correctly.</p>
  
  <pre><code>&lt;form action="formsubmit.php?team_id=6" method="post"&gt;
      &lt;label&gt; &lt;br /&gt;What did they say?: &lt;br /&gt;
          &lt;textarea name="quotetext" rows="10" cols="26"&gt;&lt;/textarea&gt;&lt;/label&gt;
      &lt;select name='name'&gt;
          &lt;?php
              while ($temp = mysql_fetch_assoc($query)) {
                  echo "&lt;option value=" . $temp['id'] . "&gt;".htmlspecialchars($temp['lastname']) . ", " . htmlspecialchars($temp['firstname']) . "&lt;/option&gt;";
              }
          ?&gt;
      &lt;/select&gt;
      &lt;input type="submit" value="Submit!" /&gt;
  &lt;/form&gt;
  </code></pre>
  
  <p>What I'm attempting to do is pass the person's ID into along to the file formsubmit.php which is called on the submission of the form. When I use $die($sql) on my database query in formsubmit.php, the ID of the person is blank.... everything else gets passed through just fine. Here is the relevant code in formsubmit.php:</p>
  
  <pre><code>$quotetext = $_POST['quotetext'];
      $id = $_POST['id'];
      $team_id = $_GET['team_id'];
      $sql = "INSERT INTO quotes SET 
              speaker_id='$id',
              quotetext='$quotetext',
              game_id=2";
      die($sql);
  </code></pre>
  
  <p>EDIT: Fixed, credit to Michael.</p>
  
  <pre><code>&lt;select name='name'&gt;
  </code></pre>
  
  <p>Should be:</p>
  
  <pre><code>&lt;select name='id'&gt;
  </code></pre>