| function
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tr8er8
msg:3806909 | 8:31 pm on Dec 13, 2008 (gmt 0) |
hi i have a function that takes a value from a row, and converts the number to text. except im having trouble getting it to work so i came here to see if anyone can see what im doing wrong:
function userLvl($value) {
if(!$_SESSION['logged_in'])
{
$access = FALSE;
}else {
$query = mysql_query("SELECT * FROM users WHERE userid = '$value'");
$row = mysql_fetch_array($query);
$num_rows = mysql_num_rows($query);
$value = $row['accessLvl'];$userLvl = array(
"0" => "User",
"1" => "Power User",
"2" => "Moderator",
"3" => "Admin"
); $userLvl = array_values($userLvl);
return $userLvl[$value];
}
if($access==FALSE)
{
global $loginUrl;
header("Location: $loginUrl");
}
}
then i do:
echo userLvl('1');
to display it
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sdlas
msg:3806910 | 8:37 pm on Dec 13, 2008 (gmt 0) |
Mann careful
i guess your script is vulnerable for a SQL injection Exploit
filter the $value
i just done a mini pentest :D for free
sorry couldn't help with your problem
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Mahabub
msg:3807112 | 10:50 am on Dec 14, 2008 (gmt 0) |
Dear tr8er8, I didnt found any error on your function. please check database connectivity and post the error which you got. also check all the values of column accessLvl in users table >=0 and <=3
Thanks
Mahabub
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tr8er8
msg:3807239 | 5:23 pm on Dec 14, 2008 (gmt 0) |
Oops! i found the problem, in the query i needed to filter where accessLvl = '$value' not userid = '$value' :D
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