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J-Rap
02-11-2006, 05:15 PM
Hey.

I had this alert system coded.

Obviously it alerts something if its in database...


And i have this error

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/local/psa/home/vhosts/Habbo-World.co.uk/httpdocs/usersystem/templates/MDStyle/alert/alert.php on line 7


Here is the code

<?
session_start();
INCLUDE "db.php";

echo $_SESSION['username'];
$sql = mysql_query("SELECT * FROM users_users WHERE username='".$_SESSION['username']."' AND alert='0'");
$alert_check = mysql_num_rows($sql);
if($alert_check > 0){
$result = mysql_query("SELECT * FROM alert");
$row = mysql_fetch_array($result);
$alert = $row['alert'];
$alert = stripslashes($alert);
include "box.php";
$query = "UPDATE user_users SET alert='1' WHERE username='".$_SESSION['username']."'";
mysql_query($query);
}
?>

Thanks peoples

vinyl-junkie
02-11-2006, 07:44 PM
You're probably getting that error because something is wrong with that first query. Modify your script as follows and see what it shows you:

<?
session_start();
INCLUDE "db.php";

echo $_SESSION['username'];
$str_sql = ("SELECT * FROM users_users WHERE username='".$_SESSION['username']."' AND alert='0'";
echo "<br>".$str_sql."<br>";
$sql = mysql_query($str_sql) or die("Error: ". mysql_error());
$alert_check = mysql_num_rows($sql);
if($alert_check > 0){
$result = mysql_query("SELECT * FROM alert");
$row = mysql_fetch_array($result);
$alert = $row['alert'];
$alert = stripslashes($alert);
include "box.php";
$query = "UPDATE user_users SET alert='1' WHERE username='".$_SESSION['username']."'";
mysql_query($query);
}
?>

J-Rap
02-13-2006, 11:05 AM
<?
session_start();
INCLUDE "db.php";

echo $_SESSION['username'];
$str_sql = ("SELECT * FROM users_users WHERE username='".$_SESSION['username']."' AND alert='0'";
echo "<br>".$str_sql."<br>";
$sql = mysql_query($str_sql) or die("Error: ". mysql_error());
$alert_check = mysql_num_rows($sql);
if($alert_check > 0){
$result = mysql_query("SELECT * FROM alert");
$row = mysql_fetch_array($result);
$alert = $row['alert'];
$alert = stripslashes($alert);
include "box.php";
$query = "UPDATE user_users SET alert='1' WHERE username='".$_SESSION['username']."'";
mysql_query($query);
}
?>

And i got an error.

Im unsure what to do..

Error =

Parse error: syntax error, unexpected ';' in /usr/local/psa/home/vhosts/Habbo-World.co.uk/httpdocs/usersystem/templates/MDStyle/alert/alert.php on line 6

vinyl-junkie
02-13-2006, 06:32 PM
Change this statement:

$str_sql = ("SELECT * FROM users_users WHERE username='".$_SESSION['username']."' AND alert='0'";

to this:

$str_sql = "SELECT * FROM users_users WHERE username='".$_SESSION['username']."' AND alert='0'";