Search Query don't show results - php

Question

I have a search engine but the problem is the search engine don't show search results, the connections to the database are correct here is the code:
PHP :
if (isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i", "", $searchq);
$query = mysql_query("SELECT * FROM tReference WHERE sReference LIKE '%$searchq%' OR sSearch LIKE '%$searchq%' OR sSort LIKE '%$searchq%'") or die("La Recherche est impossible");
$count = mysql_num_rows($query);
if ($count == 0) {
$output = "Aucun Résultat Pour Cette Recherche!";
//English Trans : No results have been Found!
}
else
{
$sReference = $count['sReference'];
$output.= '<div><ul><li><a target="_blanc" href="refrences.php?reference=' . $sReference . '" title="' . $sReference . '">' . $sReference . '</a></li></ul></div>';
}
}
HTML:
<div id="searchd">
<div class="searchc">
<form action="reference.php" method="post">
<input type="text" name="search" palceholder="Recherhcer...">
<input type="submit" Value="Ok">
<div class="clear"></div>
</form>
</div>
<div id="output">
<?php echo ($output); ?>
</div>
</div>
And Thanks to All in advance!

Answer 1

You were not actually processing the result of your query.
You have to fetch each row from the returned result set and then use the fields returned in your html output.
Try this:-
if (isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i", "", $searchq);
$query = mysql_query(
"SELECT * FROM tReference
WHERE sReference LIKE '%$searchq%' OR
sSearch LIKE '%$searchq%' OR
sSort LIKE '%$searchq%'") or die("La Recherche est impossible");
if (mysql_num_rows($query) == 0) {
$output = "Aucun Résultat Pour Cette Recherche!";
//English Trans : No results have been Found!
} else {
$output .= '<div><ul>';
while ( $row = mysql_fetch_object($query) ) {
$output .= '<li>';
$output .= '<a target="_blanc" href="refrences.php?reference=' . $row->sReference . '" ';
$output .= '" title="' . $row->sReference . '">' . $row->sReference . '</a>';
$output .= '</li>';
}
$output .= '</ul></div>';
}
}

Answer 2

It's done Guys, Thanks to all here's the solution:
if (isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace("#[^0-9a-z]#i", "", $searchq);
$query = mysql_query(
"SELECT * FROM tReference
WHERE sReference LIKE '%$searchq%' OR
sSearch LIKE '%$searchq%' OR
sSort LIKE '%$searchq%'") or die("La Recherche est impossible");
if (mysql_num_rows($query) == 0) {
$output = "Aucun Résultat Pour Cette Recherche!";
} else {
$output .= '<div><ul>';
while ( $row = mysql_fetch_assoc($query) ) {
$sReference = $row['sReference'];
$output .= '<li>';
$output .= '<a target="_blanc" href="refrences.php?reference="' . $sReference . '" ';
$output .= '" title="' . $sReference . '">' . $sReference . '</a>';
$output .= '</li>';
}
$output .= '</ul></div>';
}
}
Thanks to #RiggsFolly, and Thanks to All of you.

Answer 3

Your code makes no sense. First I recommend that you use Object oriented PHP on a separate secure page. Then your code should look more like:
//secure.php
<?php
function db(){
return new mysqli('host', 'user', 'password', 'database');
}
?>
//reference.php
include 'secure.php'; $output = '';
if(isset($_POST['sub'])){
$db = db(); $searchq = $_POST['search'];
// What is this for? --->>> $searchq = preg_replace('/[^0-9a-z]/i', "", $searchq);
$query = $db->query("SELECT * FROM tReference WHERE sReference LIKE '%$searchq%' || sSearch LIKE '%$searchq%' || sSort LIKE '%$searchq%'");
if($query->num_rows < 1){
die('Aucun Résultat Pour Cette Recherche!');
}
else{
while($row = $query->fetch_assoc()){
$sref = $row['sReference'];
$output.= "<li><a target='_blank' href='refrences.php?reference=$sref' title='$sref '>$sref</a></li>";
}
$query->free(); $db->close();
}
<div id='searchd'>
<div class='searchc'>
<form action='reference.php' method='post'>
<input type='text' name='search' placeholder='Recherhcer...' />
<input type='submit' value='Ok' name='sub' />
<div class='clear'></div>
</form>
</div>
<ul id='output'><?php echo $output; ?></ul>
</div>