penet.biz

Hi, I have this code: <?PHP require ("connect.php"); //$db_handle = mysql_connect($server, $user_name, $password); =>$connect $db_found = mysql_select_db($_mysql_db, $connect); if ($db_found) { $SQL = "SELECT * FROM cheats"; $result = mysql_query($SQL); $i=0; while ($db_field = mysql_fetch_assoc($result)) { if($i%2===0) { $mesaj="<li><a href=\"{$db_field['Image']}\" class=\"pirobox\" title=\"{$db_field['Titlu']}\"><img src=\"{$db_field['Image']}\"/></a><p>{$db_field['Content']}</p><div class=\"button\"><pre><a href=\"#\">Read more</a> <a href=\"{$db_field['Download']}\">Download</a></pre></div></li>"; print ("$mesaj"); …

I need that when i click a button to open a popup text box(something like the box used for images) , and i need to use it whith php , to put it in php code so i can put it to show me a php string that i extract …

I have this error :Parse error: syntax error, unexpected '$db_field' (T_VARIABLE) in C:\Program Files (x86)\EasyPHP-12.1\www\penet.hi2.ro\cheats.php on line 110 And the code is : <?PHP require ("connect.php"); //$db_handle = mysql_connect($server, $user_name, $password); =>$connect $db_found = mysql_select_db($_mysql_db, $connect); if ($db_found) { $SQL = "SELECT * FROM cheats"; $result = mysql_query($SQL); while ($db_field …