[PHP]Dziwny blad mysql_fetch_array Opcje

[karer]

Napisałem sobie skrypt:

[PHP] pobierz, plaintext 
<?php
error_reporting(E_ALL); 
ini_set('display_errors','1');
include('baza.php');
$id = $_GET['user'];
$user = $_SESSION['zalogowany'];
$name = mysql_fetch_array(mysql_query("SELECT `nazwa` FROM users WHERE user='$user' "));
$rank = mysql_fetch_array(mysql_query("SELECT `ranga `FROM users WHERE user='$user' "));
$hp = mysql_fetch_array(mysql_query("SELECT `hp` FROM users WHERE user='$user' "));
include('baza.php');
echo "Nazwa: " . $name . "<br>";
echo "Ranga: " . $rank . "<br>";
echo "HP" . $hp . "<br>";
?>
[PHP] pobierz, plaintext 

ale kiedy wejde na strone to mam errora:

[PHP] pobierz, plaintext 
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/forumtop/public_html/herowars/pmenu.php on line 5
 
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/forumtop/public_html/herowars/pmenu.php on line 6
[PHP] pobierz, plaintext 

Ten post edytował karer 20.08.2010, 10:04:45

[karer]

ok full skrypt:

[PHP] pobierz, plaintext 
<?php
include('baza.php');
$id = $_GET['user'];
$user = $_SESSION['zalogowany'];
$nazwa = mysql_fetch_array(mysql_query("SELECT `nazwa` FROM users WHERE user='$user' "));
$rank = mysql_fetch_array(mysql_query("SELECT `ranga` FROM users WHERE user='$user' "));
$hp = mysql_fetch_array(mysql_query("SELECT `hp` FROM users WHERE user='$user' "));
$name1 = mysql_query($nazwa) or die(mysql_error());
$rank1 = mysql_query($rank) or die(mysql_error());
$hp1 = mysql_query($hp) or die(mysql_error());
echo "Nazwa: " . $name1 . "<br>";
echo "Ranga: " . $rank1 . "<br>";
echo "HP" . $hp1 . "<br>";
?>
[PHP] pobierz, plaintext 

i error:

[PHP] pobierz, plaintext 
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/forumtop/public_html/herowars/pmenu.php on line 5
Query was empty
[PHP] pobierz, plaintext 

Ten post edytował karer 20.08.2010, 10:19:34