第二届vecctf-wp

misc

zip-guessing

打开压缩包发现提示password=.......py

进行6位密码爆破，得到密码3a1b6c

解压得到flag

flag{fa8d910600a5467d9d2f433f2bba4c80}

Spiderman

修改高度，发现字符串README：HAHAyoufoundIT

binwalk分离出一个带密码的压缩包

压缩包中含有一个名为README.txt的文件
将HAHAyoufoundIT写入README.txt并压缩成zip文件后发现两个README.txt的crc一致

采用明文爆破出密码!g^c*0

解压后得到一张png

stegsolve分析图片，发现0通道中藏有flag

flag{Z1p_1maGE_lSB_1nterestiNG}

crypto

baby rsa

因为两组数据的N相同，尝试共模攻击

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import random
import gmpy2
n,e1 = (18278250860121129714191417969518671565676609370438974713591911937173608105906651505337301762681939894104610580889645217298111784819358890536088923234106589063389221021186473430580352803412280849862104635302801082728967968107748121682339447067731887326087497520544915737621714860811217331095698945949486830528216492874665673677441699322439459330337628648399309373351992766984612437555092804129107438390294123179182945321411199463748963590901018595492869891970965012682196059824464339830536600021131729277503225240822584625293463254992623570796895110736563514265569133338477447460712899749955523316083254492005199917987, 76831)
c1 = 14068017239200053057485191549313171235040341952680207740172042657952395329165768631549600187463451884623632792268314205814005668826033658861436424617991125160661381291003863394279707111450435808077534541510943745285958554921154542813613390731768463465972495263609910077893559813685204864361223917445770159726623708768996447642959844639282015350260298045019617026128700143368357847147111693153254268063490280666808714031170237337721534721243540077707555520501582047389390048200750092516532861791678414405037042484148584170396597267139826760931348123330503380212299827883145622326471261247290265669559897644089748984204
n,e2 = (18278250860121129714191417969518671565676609370438974713591911937173608105906651505337301762681939894104610580889645217298111784819358890536088923234106589063389221021186473430580352803412280849862104635302801082728967968107748121682339447067731887326087497520544915737621714860811217331095698945949486830528216492874665673677441699322439459330337628648399309373351992766984612437555092804129107438390294123179182945321411199463748963590901018595492869891970965012682196059824464339830536600021131729277503225240822584625293463254992623570796895110736563514265569133338477447460712899749955523316083254492005199917987, 65537)
c2 = 17120091052055468753592285548832983539343349280048150965811042901415659935929189229298761316319784219958366772027986939749507760509991462838561325114714990126772560070483939658678703332135866293838146121312972251030190756953273311362066029287803984179301942229948301600814847050270084193184210232673652417884727838290706970298245205101865234726242081382680212621541249662659284218079172180588507736208257188130972936994297239579885142233896240153633237886314935309710560434501987121687537279749951910677771351212584805615709915168729977338949858374823750316083595198620089092891606394712323216539675547753350603350269
s=gmpy2.gcdext(e1,e2)
s1=s[1]
s2=s[2]
if s1<0:
    s1=-s1
    c1=gmpy2.invert(c1,n)
else:
    s2=-s2
    c2=gmpy2.invert(c2,n)

c1=gmpy2.powmod(c1,s1,n)
c2=gmpy2.powmod(c2,s2,n)
c=(c1*c2)%n
print(bytes.fromhex(hex(c)[2:]))

flag{423c35691377ea18d35b97b7b6f13590}

web

ezfilter

抓包，在密码处尝试sql注入，回显了mysql的报错

fuzz后发现过滤了空格，可以用/**/代替空格

sqlmap跑一下得到flag
sqlmap -r 1.txt --batch --tamper "space2comment.py" --dump

flag{30b08224ed436c8c690ae623a89466f3}

upload

在源码中发现hint: TUhnd2VEQjRiRzluYVc0dWNHaHc=

两层base64得到0x0x0xlogin.php

通过弱口令爆破进入后台 admin,admin123456

在项目管理处发现一个文件上传点

通过将文件名置空，报错得到上传路径uploaded

fuzz后发现允许上传.htaccess

再上传一个图片马即可

蚁剑连接得到flag

flag{063a5bda0b5a2a5d7b700d989f73902b}

WEB2

查看源码：/index.php?source

关键代码如下

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if (isset($_POST['session_id'])) {
    $query = "SELECT * FROM secrets WHERE session_id = '" . $_POST['session_id'] . "'";
    $result = $conn->query($query);
} else {
    $query = "SELECT * FROM secrets WHERE session_id = '" . session_id() . "'";
    $result = $conn->query($query);
}

where后面为true就行了

payload：1' or '1'='1

EzCas

Apereo Cas反序列化

题目不出网，公开的普通exp没有回显
在github上找到一个带回显的项目

生成payload

构造命令

Mobile

findme

模拟器先安装一下，有个查询按钮

apk文件，使用jadx分析

先全局搜索找到onClick函数，查看点击查询按钮时执行的代码

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public class OooO0O0 implements View.OnClickListener {

    /* renamed from: OooO0o0  reason: collision with root package name */
    public final /* synthetic */ EditText f1853OooO0o0;

    public OooO0O0(EditText editText) {
        this.f1853OooO0o0 = editText;
    }

    public void onClick(View v) {
        try {
            LoginActivity.this.OooOo0o.OooO0Oo(this.f1853OooO0o0.getText().toString());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

发现调用了LoginActivity.this.OooOo0o.OooO0Oo，查看函数声明

发现存在数据库

getWritableDatabase函数用来连接数据库
OooO0OO函数用来获取数据库密码
this.f1221OooO0o为OooO0OO的对象

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public void OooO0O0(String t) {
    this.OooO0oO = t;
    SQLiteDatabase.loadLibs(this.f1222OooO0o0);
    this.f1221OooO0o = new OooO0OO(this.f1222OooO0o0);
    String OooO0OO2 = OooO0OO();
    this.f1220OooO0Oo = OooO0OO2;
    this.f1219OooO0OO = this.f1221OooO0o.getWritableDatabase(OooO0OO2);
    new OooO0O0();
    for (int g = 0; g < 100; g++) {
        ContentValues values = new ContentValues();
        values.put("flag", OooO0O0.OooO00o("" + g));
        values.put("num", Integer.valueOf(g));
        this.f1219OooO0OO.insert("test", null, values);
    }
}


public void OooO0Oo(String i) {
    SQLiteDatabase.loadLibs(this.f1222OooO0o0);
    this.f1221OooO0o = new OooO0OO(this.f1222OooO0o0);
    String OooO0OO2 = OooO0OO();
    this.f1220OooO0Oo = OooO0OO2;
    this.f1219OooO0OO = this.f1221OooO0o.getWritableDatabase(OooO0OO2);
    Cursor cursor = this.f1219OooO0OO.rawQuery("select * from test where num = '" + i + "'", null);
    this.OooO0oo = "";
    if (cursor == null || !cursor.moveToFirst()) {
        Toast.makeText(this.f1222OooO0o0, this.OooO0oo, 0).show();
        cursor.close();
        this.f1219OooO0OO.close();
    }
    do {
        this.OooO0oo += cursor.getString(cursor.getColumnIndex("flag")) + "\n";
    } while (cursor.moveToNext());
    Toast.makeText(this.f1222OooO0o0, this.OooO0oo, 0).show();
    cursor.close();
    this.f1219OooO0OO.close();
}

通过OooO0OO类的声明获得数据库文件的名称data.db

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public class OooO0OO extends SQLiteOpenHelper {
    public OooO0OO(Context context) {
        super(context, "data.db", null, 1);
    }

    @Override // net.sqlcipher.database.SQLiteOpenHelper
    public void onCreate(SQLiteDatabase sqLiteDatabase) {
        sqLiteDatabase.execSQL("create table test(flag text, num text)");
    }

    @Override // net.sqlcipher.database.SQLiteOpenHelper
    public void onUpgrade(SQLiteDatabase sqLiteDatabase, int i, int i1) {
        onCreate(sqLiteDatabase);
    }
}

改apk的后缀名然后解压文件，找到data.db文件

使用DB4S打开db文件，发现需要密码

获取OooO0OO函数的代码，发现先对this.OooO0oO中储存的数据进行base64解密
然后将返回值作为参数执行OooO00o.OooO00o函数
并将OooO00o.OooO00o函数的返回值作为参数调用OooO00o

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public String OooO0OO() {
    return OooO00o(new String(OooO00o.OooO00o(Base64.decode(this.OooO0oO, 0))));
}

对this.OooO0oO查找用例，找到获取资源的代码

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public String Oooo0o0() {
    String flag = getSharedPreferences("secret", 0).getString("flag", null);
    Log.i("secret", "读取信息");
    Log.i("secret", "flag:" + flag);
    return flag;
}

搜索资源secret，获取到加密后的密码mkrPETW43g1796lzN2dtiC5qgsC+ybJIWHR8GEzhsI4=

查看OooO00o.OooO00o函数的代码，发现是aes解密，密钥通过OooO0O0函数获取

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public class OooO00o {
    public static byte[] OooO00o(byte[] plainText) {
        SecretKeySpec secretKey = new SecretKeySpec(OooO0O0("xGP_UACWNcWO]WNP".getBytes(), "123456".getBytes()), "AES");
        Cipher cipher = Cipher.getInstance("AES");
        cipher.init(2, secretKey);
        return cipher.doFinal(plainText);
    }

    public static byte[] OooO0O0(byte[] data, byte[] key) {
        if (data == null || data.length == 0 || key == null || key.length == 0) {
            return data;
        }
        byte[] result = new byte[data.length];
        for (int i = 0; i < data.length; i++) {
            result[i] = (byte) ((data[i] ^ key[i % key.length]) ^ (i & 255));
        }
        return result;
    }
}

通过脚本得到密钥Itahdrtbu^hr`hsk

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a=b'xGP_UACWNcWO]WNP'
b=b'123456'
e=''
for i in range(len(a)):
    e+=chr(a[i]^b[i%len(b)]^i)

print(e)

aes解密得到OooO00o函数的参数2>=9:92m;<in=?3n

将f1218OooO0O0变量的所有成员与参数的每一位异或FECanfOI()*&<MNCXZPKL

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static {
    Charset forName = Charset.forName("UTF-8");
    OooO00o = forName;
    f1218OooO0O0 = "FECanfOI()*&<MNCXZPKL".getBytes(forName);
}

public static String OooO00o(String enc) {
    byte[] b = enc.getBytes(OooO00o);
    int size = b.length;
    for (int i = 0; i < size; i++) {
        for (byte keyBytes0 : f1218OooO0O0) {
            b[i] = (byte) (b[i] ^ keyBytes0);
        }
    }
    return new String(b);
}

通过脚本解出数据库的密码9562129f07be648e

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key=b"FECanfOI()*&<MNCXZPKL"
data="2>=9:92m;<in=?3n"
flag=''
for i in data:
    c=i
    for j in key:
        c=c^j
    flag+=chr(c)
print(flag)

进入数据库后发现表test中有一条特别的数据（与其他数据长度不一样），fee0b212f413f8a0

根据长度猜测flag是密码加最后一条数据

flag{9562129f07be648efee0b212f413f8a0}

wp
#wp #2022竞赛 #vec
第二届vecctf-wp
https://www.dr0n.top/posts/e6acd4f1/
作者
dr0n
发布于
2022年5月21日
更新于
2025年7月27日
许可协议