php - 搜索查询不显示结果

Question

我有一个搜索引擎，但问题是搜索引擎不显示搜索结果，与数据库的连接是正确的，这里是代码:

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>

提前感谢大家!

Answer 1

您实际上并未处理查询结果。您必须从返回的结果集中获取每一行，然后使用 html 输出中返回的字段。

试试这个:-

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>';
    }

}