استفسار من ذوي الخبرة في برمجات ال PHP
السلام عليكم
لدي استفسار لمذا لا يعمل الكود التالي????
كود PHP:
$section_name=$_GET['name'];
$change_db= "sound_link";
$change_status = "1";
$change_order = "link_id";
$change_search_id = "link_id";
$change_search_title = "name";
$change_search_cat = "cat";
$change_search_cat_url = "cat";
$like_tables= "linkname"; //('body' OR 'body_html')
$get_query=$_POST[query];
$search_query=mysql_query("SELECT * FROM ".$change_db." WHERE ".$like_tables." LIKE '%$get_query%') AND status='$change_status' ORDER BY $change_order DESC");
echo "<table class=\"songs_search_results\">";
$i=0;
while($search_row=mysql_fetch_array($search_query)){
$result_id=$search_row['$change_search_id'];
$result_title=$search_row['$change_search_title'];
$result_cat=$search_row['$change_search_cat'];
$i++;
echo "<tr><td align=\"center\" width=25>$i</td>
<td align=\"right\"><a class=newshead href=\"sections.php?name=$section_name&file=details&$change_search_cat_url=$result_cat&id=$result_id\">$result_title</td>
</tr>";
}
echo "</table";
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
الخلل في سطر ال $search_query
