Parašė Grold· 2011 Bal. 10 10:04:40
#6
Va si koda dedu:
$dv = mysql_query("SELECT dovana from dovanos WHERE kam='".$user_data['user_id']."' ORDER BY laikas DESC LIMIT 0,1");
while ($q = mysql_fetch_array($dv))
{
$dovana = $q['dovana'];
}
echo "<img src='".IMAGES."dovanos/$dovana.png'>";
Galbut gali duoti pilnai kaip turi atrodyti?
O cia klauset koki error tai:
Notice: Undefined variable: dovana in /home/sampoa/domains/manodomenas.lt/public_html/profile.php on line 51
Redagavo Grold· 2011 Bal. 10 10:04:48