Drop down menu error-display error - drop-down-menu

Question

i'm in the middle of creating a website and i want to creat a dynamic dropdown menu.The problem is that when i load the entries from the database only the first entry is displayed correctly , the other entries in the menu are displayed as code. I'm using Xamp with PHP 5.3 and HeidiSql.This is my code :
<div id="1" class="mega solidblocktheme">
<div class="column">
<ul>
<li><a href="<?php include('../conect.php'); $result=mysql_query("SELECT * FROM tours");
while($data=mysql_fetch_row($result))
{ echo $data[2]; ?>"> <?php echo $data[1];
echo "</a></li>"; }
mysql_close($con);?>
</ul>
</div>
</div>

Answer 1

Your loop seems to be in the wrong place. Try something like this:
<ul>
<?php include('../conect.php');
$result=mysql_query("SELECT * FROM tours");
while($data=mysql_fetch_row($result))
{
echo '<li><a href="'.$data[2].'">'.$data[1];
echo "</a></li>";
}
mysql_close($con);?>
</ul>

Answer 2

You should not do it this way .
while($data=mysql_fetch_row($result))
{ echo $data[2]; ?>">
echo ""; }
For every loop , you are just adding content and close tags </a> and </li>
But you are not adding start tags for them .
What you can do is , first store your data result in a string
include('../conect.php');
$result=mysql_query("SELECT * FROM tours");
$data=mysql_fetch_row($result);
mysql_close($con);
$list = "";
foreach($data as $data_key) {
$list.= '<li> '.$data[1].' </li>';
}
So now you got your list
After that comes your HTML , just echo your built string
<div id="1" class="mega solidblocktheme">
<div class="column">
<ul>
<?php
echo $list;
?>
</ul>
</div>
</div>