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IT Security and Insecurity Portal |
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mysql injection help |
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Posted: Wed May 04, 2005 10:14 pm |
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james |
Beginner |
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Joined: May 05, 2005 |
Posts: 4 |
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I found this coming while on asite and i was wounder how cna i inject a code into it
if i can update my account
Code: | andom snippets from a while ago:
$connection2=safeConnect($host,$dbusername,$dbpassword); } if(($userid == 2236774) && ($ip != '70.88.194.78')) $user['admin']==3;
//...
$dbusername="remote"; $dbpassword="99bignoob";
//...
if($user['money']>50000000000) { dbresult("UPDATE users SET suspended=1, suspend_reason='Auto-suspended: Money Glitch Creator' WHERE id=$userid"); err("This account looks as though it has created a money glitch and we are investigating.",1); } */
//...
if($user['points']>20000) { //dbresult("UPDATE users SET suspended=1, suspend_reason='Auto-suspended: Money Glitch Creator' WHERE id=$userid"); dbresult("UPDATE users SET suspended=1, suspend_reason='Account suspended for point glitch investigation. Please leave your AIM screenname in your rebuttle.' WHERE id=$userid"); err("This account looks as though it has created a point glitch and we are investigating.",1); } if($user['points']<0) { $reason = $user['points'] . " points on script " . $script_name; suspenduser($userid, $reason); err("Negative points.",1); } $isadventure = 0; $israid = 0;
//...
if($ontest && !in_array($userid, $validusers)) { if(($pass != "shazbot") && ($batman<>1)) //Used for backend login as user { include "includes/header.inc"; err("Invalid server request! Please use www.******.com", 1); } }
//...
MySQL Configuration $www1 = "66.45.249.34"; $www2 = "66.45.249.35"; $database_connections[1][] = $www1; $database_connections[1][] = $www2; $torax1 = "66.115.249.36"; $torax2 = "66.45.245.37"; $database_connections[2][] = $torax1; $database_connections[2][] = $torax2; $fabar1 = "66.25.249.41"; $fabar2 = "66.45.223.42"; $database_connections[3][] = $fabar1; $database_connections[3][] = $fabar2; $cl1 = "66.45.112.38"; $chatdb = "chat"; $dbusername="rem0te"; $dbpassword="99bignoob"; $sname = $_SERVER["SCRIPT_NAME"]; for($x=0; $x |
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Posted: Wed May 04, 2005 10:18 pm |
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james |
Beginner |
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Joined: May 05, 2005 |
Posts: 4 |
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and found this Code: | Error in query: INSERT INTO newforum (id, userid, username, subject, day, month, year, message, front, pinned, crewid, category) VALUES (\'0\', \'7706\', \'James?\', \'test\', \'04\', \'05\', \'2005\', \'test\'\', \'yes\', \'no\', \'37283\', \'\') You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near \'yes\', \'no\', \'37283\', \'\')\' at line 1 | when ever i use a ' thing ill get this error and if i dont it post It a lil ghetto forum the sites uses |
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