PHP display database using a search result - php

Question

I am currently trying to create a search menu, and display it's result in a form of table. I managed to get the search result only as a string, but unable to use it to display a full row in the table.
Here is my code :
- Creating the Form :
<form action = "profile.php" method = "post">
<input type="text" name="search" placeholder = "Search">
<input type="submit" value = "SEARCH"><br>
</form>
This is the php to get search result
if(isset($_POST['search']))
{
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$query = mysql_query("SELECT * FROM identity WHERE fullname LIKE '%$searchq%'") or die("Could not Find!");
$count = mysql_num_rows($query);
if($count == 0)
{
$output = 'There is no result';
}
else
{
while($row = mysql_fetch_array($query))
{
$fullname = $row['fullname'];
$output .= '<div> '.$fullname.' </div>';
$records = mysql_query("SELECT * FROM identitas_pengadu WHERE fullname='$output'");
}
}
}
This is the code I used to display the result in a database :
while($identity = mysql_fetch_assoc($records))
{
echo "<tr>";
echo "<td>".$identity['fullname']."</td>";
echo "<td>".$identity['nip']."</td>";
echo "<td>".$identity['email']."</td>";
echo "<td>".$identity['position']."</td>";
echo "<td>".$identity['status']."</td>";
echo "<td>EDIT</td>";
echo "</tr>";
}
It works fine if the value of $records does not use the WHERE function. Please help.
Thank You