php if inside if execute sql statement - php

Question

the form already has an existing submit button which you can select difficulty(1st combo) and location(2nd combo) and then i submit this to display a video and a text field
under the field i have an addtional drop down box which is yes or no
then inside this code i want to submit a separate value using an if inisde an if it works fine on the 1st difficulty and location in the lists but nothing else and i cant see what im doing wrong a new pair of eyes would be great PS i will be upgrading this to PDO eventually
<?php
if(strpos($drop, 'norm') !== false && $ruavalue == 0)
{
Echo "RUA: ";
$RUAresult = mysql_query("SELECT Answer FROM options");
echo "<select name='ruacombo'>";
while($RUArow = mysql_fetch_assoc($RUAresult))
echo "<option value = '".$RUArow['Answer']."'>".$RUArow['Answer']."</option>";
echo "</select>";
echo '<form action="" method="post">';
echo "<input name=\"boss\" type=hidden value=".$_POST['tier_two'].">";
echo "<input name=\"main\" type=hidden value=".$_COOKIE['ID_my_site'].">";
echo '<input type="submit" name="ruasubmit" value="RUA!" />';
echo '</form>';
if (isset($_POST['ruasubmit']))
{
$ruaboss = $_POST['boss'];
$ruauser = $_POST['main'];
$ruasql = "UPDATE `RUASEXCELL` SET `$ruaboss`=1 WHERE Username = '$ruauser'";
$add_rua = mysql_query($ruasql);
}

Answer 1

In your while loop, you need to access the associative array keys as a string like so:
while($RUArow = mysql_fetch_assoc($RUAresult))
echo "<option value = '".$RUArow['Answer']."'>".$RUArow['Answer']."</option>";
Also, please look into using mysql_real_escape_string() to prevent SQL injection. In its current state, this script, and your database can be blown to pieces by a stray quote.