[ tandrli @ 25.09.2014. 13:00 ] @
poz imam potrebu da izvucem variablu ulogovanom korisiku recimo "lucky232" koja je upisana u istoj tabeli ako pogledate na dnu Code: echo "zzzzz $rrr"; tamo ispisuje ime usera tj. zzzzz lucky232 ako zamjenim $rrr sa userom .. nakon pretrage on printa trazenu variablu .. medjutim sve dok imam (user_name ="$rrr") pretraga nece ni da cuje Code: $sql = 'SELECT prskola FROM users WHERE user_name ="$rrr"'; //nece da radi za $sql = 'SELECT prskola FROM users WHERE user_name ="lucky232"'; //radi Code: if ($login->isUserLoggedIn() == true) { include("views/logged_in.php"); include("br.html"); $_SESSIO = $_SESSION['user_name']; $rrr="$_SESSIO"; $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = '111'; $conn = mysql_connect($dbhost, $dbuser, $dbpass); if(! $conn ){ die('Could not connect: ' . mysql_error()); } $sql = 'SELECT prskola FROM users WHERE user_name ="$rrr"'; mysql_select_db('login'); $retval = mysql_query( $sql, $conn ); if(! $retval ){ die('Could not get data: ' . mysql_error()); } while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) { echo "to je to {$row['prskola']} <br> ". "--------------------------------<br>"; } mysql_close($conn); echo "zzzzz $rrr"; } else { include("views/not_logged_in.php"); } pa ako ko moze sa me uputi .. đe je zapelo thnx |